Refinements of the Erdös-Mordell inequality, Barrow’s inequality, and Oppenheim’s inequality

Abstract

In this paper, we present some new refinements of the Erdös-Mordell inequality, Barrow’s inequality, and Oppenheim’s inequality. Based on verification by computer, several related interesting conjectures are put forward.

1 Introduction

For any interior point P of the triangle $ABC$, let $R_{1}$, $R_{2}$, $R_{3}$ denote the distances from P to the vertices A, B, C, and let $r_{1}$, $r_{2}$, $r_{3}$ denote the distances from P to sides BC, CA, AB, respectively. In addition, we denote the cyclic sums over the triples $(R_{1}, R_{2}, R_{3})$ and $(r_{1}, r_{2}, r_{3})$ by ∑. Then the well-known Erdös-Mordell inequality states that

$$ \sum R_{1}\ge2\sum r_{1}, $$

(1.1)

with equality if and only if $\triangle ABC$ is equilateral and P is its center.

Inequality (1.1) was conjectured by Erdös [1] in 1935. Mordell and Barrow [2] first proved it in 1937, and since then this inequality is known as the Erdös-Mordell inequality and has attracted the attention of many mathematicians who offered various new proofs, generalizations, variations, sharpness, and conjectures (see [3–33] and the references therein).

In [2], Barrow indeed proved the following stronger version of (1.1):

$$ \sum R_{1}\ge2\sum w_{1}, $$

(1.2)

where $w_{1}$, $w_{2}$, $w_{3}$ are the lengths of the internal bisectors of $\angle BPC$, $\angle CPA$, $\angle APB$, respectively. For generalizations, sharpness, and extensions of Barrow’s inequality, see [3, 4, 6, 10, 25, 31].

In 1961, Oppenheim [9] applied geometrical transformations to study Erdös-Mordell inequality and other inequalities connecting the segments $R_{1}$, $R_{2}$, $R_{3}$, $r_{1}$, $r_{2}$, $r_{3}$. At the end of this paper, he conjectured that the following inequality holds:

$$ \sum R_{2}R_{3}\ge\sum (r_{3}+r_{1}) (r_{1}+r_{2}). $$

(1.3)

Oppenheim [10] tried to prove this inequality in the same year. However, the author of this paper pointed out that there exist faults in Oppenheim’s proof in a recent paper [28]. We also presented a new method to prove the above inequality in the paper. In fact, we proved the following refinement of (1.3):

$$\begin{aligned} \sum R_{2}R_{3}\ge\sum h_{a}r_{1}+\sum r_{2}r_{3}\ge \sum(r_{3}+r_{1}) (r_{1}+r_{2}), \end{aligned}$$

(1.4)

where $h_{a}$, $h_{b}$, $h_{c}$ are the corresponding altitudes of $\triangle ABC$ and $\sum h_{a}r_{1}=h_{a}r_{1}+h_{b}r_{2}+h_{c}r_{3}$.

The main purpose of this paper is to establish some new refinements of the Erdös-Mordell inequality, Barrow’s inequality (1.2) and Oppenheim’s inequality (1.3). We also propose some closely related interesting conjectures.

The remainder of this paper is organized as follows. In the next section, we first establish a new inequality involving an interior point of a triangle. Then we use this inequality, the Erdös-Mordell inequality and another well-known inequality to deduce a refinement of the Oppenheim inequality. In Section 3, we present a refinement with one parameter for Barrow’s inequality (1.2). In Sections 4 and 5, some new refinements of the Erdös-Mordell inequality are established. Finally, in Section 6, we propose some interesting related conjectures as open problems.

In the following, we shall unceasingly use the above symbols. In addition, we also denote the lengths of the sides BC, CA, AB of the triangle $ABC$ by a, b, c, respectively, and denote the corresponding medians of the triangle $ABC$ by $m_{a}$, $m_{b}$, $m_{c}$. We denote cyclic sums over the triples $(a, b, c)$ (including subscripts), $(x, y, z)$, $(u, v, w)$, $(r_{1}, r_{2}, r_{3})$, $(R_{1}, R_{2}, R_{3})$, and $(A, B, C)$ by ∑, such as

$$\begin{aligned}& \sum a(b-c)^{2}r_{1}^{2}=a(b-c)^{2}r_{1}^{2}+b(c-a)^{2}r_{2}^{2}+c(a-b)^{2}r_{3}^{2},\\& \sum w_{1}(R_{2}+R_{3})=w_{1}(R_{2}+R_{3})+w_{2}(R_{3}+R_{1})+w_{3}(R_{1}+R_{2}),\\& \sum u(wy-vz)^{2}= u(wy-vz)^{2}+v(uz-wx)^{2}+w(vx-uy)^{2},\\& \sum(m_{b}+m_{c}) (r_{1}+R_{1})=(m_{b}+m_{c}) (r_{1}+R_{1})+(m_{c}+m_{a}) (r_{2}+R_{2})+(m_{a}+m_{b}) (r_{3}+R_{3}). \end{aligned}$$

2 A new refinement of Oppenheim’s inequality

In this section, we first establish a new geometric inequality which may be of independent interest. We shall make use of this result and its weaker form (see inequality (4.3) below) several times in the sequel.

Lemma 2.1

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{2}R_{3}\ge\frac{1}{2} \sum(2w_{1}+w_{2}+w_{3})R_{1}, $$

(2.1)

with equality holding if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

First of all, we establish the following weighted trigonometric inequality for any triangle $ABC$:

$$ \sum\frac{yz(2x+y+z)}{y+z}\cos A\le\sum yz, $$

(2.2)

where A, B, C are the angles of $\triangle ABC$ and x, y, z are arbitrary positive numbers.

By the law of cosines, inequality (2.2) is equivalent to

$$\sum\frac{yz(2x+y+z)(b^{2}+c^{2}-a^{2})}{bc(y+z)}\le2\sum yz, $$

i.e.,

$$\begin{aligned} &2abc(y+z) (z+x) (x+y)\sum yz \\ &\quad{}-\sum yz(z+x) (x+y) (2x+y+z)a\bigl(b^{2}+c^{2}-a^{2} \bigr)\ge0. \end{aligned}$$

(2.3)

Let $b+c-a=2u$, $c+a-b=2v$, $a+b-c=2w$, then $a=v+w$, $b=w+u$, $c=u+v$, where $u, v, w>0$ by the triangle inequality. Thus, we see that inequality (2.3) is equivalent to the following algebraic inequality:

$$\begin{aligned} &2(v+w) (w+u) (u+v) (y+z) (z+x) (x+y)\sum yz \\ &\qquad{}-\sum yz(z+x) (x+y) (2x+y+z) (v+w) \bigl[(w+u)^{2}+(u+v)^{2}-(v+w)^{2} \bigr] \\ &\quad\ge0. \end{aligned}$$

(2.4)

Expanding and rearranging gives the equivalent inequality:

$$ 4\sum x\sum ux^{2}(vy-wz)^{2}+4xyz \sum u(wy-vz)^{2}\ge0, $$

(2.5)

which is obviously true for the positive numbers x, y, z, u, v, w. From (2.5), it is easily seen that the equality in (2.4) holds if and only if $x=y=z$ and $u=v=w$. We further conclude that equality in (2.2) holds if and only if $\triangle ABC$ is equilateral and $x=y=z$.

We now make use of inequality (2.2) to deduce geometric inequality (2.1). Let $\angle BPC=2\delta_{1}$, $\angle CPA=2\delta_{2}$, and $\angle APB=2\delta_{3}$, then we have the following known identity (cf. [4], p.317):

$$ w_{1}=\frac{2R_{2}R_{3}}{R_{2}+R_{3}}\cos\delta_{1}, $$

(2.6)

and two formulas are valid for $w_{2}$ and $w_{3}$. Since $0<\delta_{1}<\pi$, $0<\delta_{2}<\pi$, $0<\delta_{3}<\pi$, and $\delta_{1}+\delta_{2}+\delta_{3}=\pi$, we see that $\delta_{1}$, $\delta_{2}$, $\delta_{3}$ are angles of a triangle. Hence, it follows from (2.2) and (2.6) that

$$\begin{aligned} \sum R_{2}R_{3} &\ge\sum(2R_{1}+R_{2}+R_{3}) \frac{R_{2}R_{3}}{R_{2}+R_{3}}\cos\delta _{1} \\ &=\frac{1}{2}\sum(2R_{1}+R_{2}+R_{3})w_{1} \\ &=\frac{1}{2}\sum(2w_{1}+w_{2}+w_{3})R_{1}, \end{aligned}$$

which proves inequality (2.1). By the equality condition of (2.2), we conclude that equality in (2.1) holds if and only if $R_{1}=R_{2}=R_{3}$ and $\delta_{1}=\delta_{2}=\delta_{3}$, i.e., $ABC$ is equilateral and P is its center. This completes the proof of Lemma 2.1. □

It is well known that the following inequality is related to the Erdös-Mordell inequality. See [9] for instance.

Lemma 2.2

For any interior point P of the triangle $ABC$, we have

$$ \sum r_{1}R_{1}\ge2\sum r_{2}r_{3}, $$

(2.7)

with equality holding if and only if P coincides with a vertex of $\triangle ABC$ or $\triangle ABC$ is equilateral and P is its center.

We now state and prove the following refinement of the Oppenheim inequality.

Theorem 2.1

For any interior point P of the triangle $ABC$, the following inequalities hold:

$$ \sum R_{2}R_{3}\ge\frac{1}{2} \sum R_{1}(2r_{1}+r_{2}+r_{3})\ge \sum(r_{3}+r_{1}) (r_{1}+r_{2}). $$

(2.8)

Equalities in (2.8) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

By Lemma 2.1 and the obvious facts that $w_{1}\ge r_{1}$, $w_{2}\ge r_{2}$, and $w_{3}\ge r_{3}$, the first inequality of (2.8) follows immediately. In addition, by the Erdös-Mordell inequality (1.1) and inequality (2.7), we have

$$\begin{aligned} &\frac{1}{2}\sum R_{1}(2r_{1}+r_{2}+r_{3})\\ &\quad=\frac{1}{2}\sum R_{1}\sum r_{1}+ \frac{1}{2}\sum r_{1}R_{1}\\ &\quad\ge \Bigl(\sum r_{1} \Bigr)^{2}+\sum r_{2}r_{3}=\sum(r_{3}+r_{1}) (r_{1}+r_{2}), \end{aligned}$$

which proves the second inequality of (2.8).

It is clear that the equality conditions in (2.8) are the same as in (1.1) and (2.1), i.e., if and only if $\triangle ABC$ is equilateral and P is its center. The proof of Theorem 2.1 is completed. □

3 A refinement of Barrow’s inequality

We first give the following compact inequality.

Lemma 3.1

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}^{2}\ge\sum w_{1}(R_{2}+R_{3}), $$

(3.1)

with equality holding if and only if $R_{1}: R_{2}: R_{3}=\sin\frac{1}{2} {\angle BPC}: \sin\frac{1}{2}{\angle CPA}: \sin\frac{1}{2}{\angle APB}$.

In fact, inequality (3.1) is a simple consequence of the following well-known weighted inequality (see [4], p.317, Theorem 12):

$$ \sum w_{1} \biggl(\frac{1}{R_{2}}+ \frac{1}{R_{3}} \biggr)yz\le\sum x^{2}, $$

(3.2)

where x, y, z are arbitrary real numbers and the equality holds if and only if $x: y: z=\sin\frac{1}{2} {\angle BPC}: \sin\frac{1}{2}{\angle CPA}: \sin\frac{1}{2}{\angle APB}$.

If we let $x=R_{1}$, $y=R_{2}$, and $z=R_{3}$ in (3.2), then inequality (3.1) follows immediately. To our surprise, inequality (3.1) has not been given in [4, 10] although it is compact.

Remark 3.1

Oppenheim [10] obtained a weighted generalization of Barrow’s inequality (1.2), which is equivalent with (3.2) when x, y, z are positive.

For Barrow’s inequality (1.2), we now give the following refinement with one parameter.

Theorem 3.1

Let k be a real number such that $0\le k\le2$, then for any interior point P of the triangle $ABC$ the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(kR_{1}+R_{2}+R_{3})(w_{1}+R_{1})}{(k+2)\sum R_{1}}\ge 2\sum w_{1}. $$

(3.3)

If $k=0$, then the first equality in (3.3) holds if and only if $R_{1}:R_{2}:R_{3}=\sin\frac{1}{2}\angle BPC:\sin\frac{1}{2}\angle CPA:\sin\frac{1}{2}\angle APB$. If $0< k\le2$, then the equalities in (3.3) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

The first inequality of (3.3) is equivalent to

$$(k+2) \Bigl(\sum R_{1} \Bigr)^{2}-2\sum (kR_{1}+R_{2}+R_{3}) (w_{1}+R_{1}) \ge0. $$

Expanding and rearranging gives

$$ 2 \Bigl[\sum R_{1}^{2}-\sum w_{1}(R_{2}+R_{3}) \Bigr]+k \Bigl[2\sum R_{2}R_{3}-2\sum w_{1}R_{1}-\sum R_{1}^{2} \Bigr]\ge0, $$

(3.4)

which is required to prove.

If $k=0$, then the above inequality becomes inequality (3.1), and the equality condition in this case is obtained from Lemma 3.1. If $2\ge k>0$, by (3.1), to prove (3.4) we need to prove that

$$k \Bigl[ \sum R_{1}^{2}-\sum w_{1}(R_{2}+R_{3}) \Bigr]+k \Bigl[2\sum R_{2}R_{3}-2\sum w_{1}R_{1}-\sum R_{1}^{2} \Bigr]\ge0, $$

i.e.,

$$ 2\sum R_{2}R_{3}-\sum (2w_{1}+w_{2}+w_{3})R_{1}\ge0, $$

(3.5)

which is equivalent with inequality (2.1) of Lemma 2.1. Hence, inequality (3.4) is proved when $0< k\le2$, while the equality condition in this case is the same as in (2.1).

The second inequality of (3.3) is equivalent to

$$\sum(kR_{1}+R_{2}+R_{3}) (w_{1}+R_{1})-(k+2)\sum w_{1}\sum R_{1}\ge0, $$

which can be rewritten as

$$ 2k \Bigl[\sum R_{1}^{2}-\sum w_{1}(R_{2}+R_{3}) \Bigr]+2\sum R_{2}R_{3}-\sum R_{1}(2w_{1}+w_{2}+w_{3}) \ge0. $$

(3.6)

By inequalities (2.1) and (3.1), we see that the above inequality holds for $k\ge0$. Also, by the equality conditions of (2.1) and (3.1), it is easy to conclud that if $k\ge0$ then the equality in (3.6) holds if and only if $\triangle ABC$ is equilateral and P is its center. This completes the proof of Theorem 3.1. □

The following particular case ($k=0$) of Theorem 3.1 is of interest.

Corollary 3.1

For any interior point P of $\triangle ABC$, we have

$$ \sum R_{1}\ge\frac{\sum(R_{2}+R_{3})(w_{1}+R_{1})}{\sum R_{1}}\ge2\sum w_{1}. $$

(3.7)

This double inequality can be regarded as an associated result of (2.1) and (3.1).

4 Refinements of the Erdös-Mordell inequality I

There are few refinements of the Erdös-Mordell inequality in the literature. In [27], the author gave the following result:

$$\begin{aligned} \sum R_{1}\ge\frac{1}{2}\sum \sqrt{a^{2}+4r_{1}^{2}}\ge\sum \biggl(\frac{c}{b}+\frac{b}{c} \biggr)r_{1}\ge2\sqrt{\sum h_{a}r_{1}}\ge2\sum r_{1}. \end{aligned}$$

(4.1)

In this section and next section, we shall give some new refinements of the Erdös-Mordell inequality. First, we point out that we have the following result, which is a counterpart of Theorem 3.1.

Theorem 4.1

Let k be real number such that $0\le k\le2$, then for any interior point P of the triangle $ABC$ the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(kR_{1}+R_{2}+R_{3})(r_{1}+R_{1})}{(k+2)\sum R_{1}}\ge 2\sum r_{1}. $$

(4.2)

Equalities in (4.2) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

Noting the fact that $w_{1}\ge r_{1}$ etc., the following two inequalities follow from (2.1) and (3.1), respectively:

$$\begin{aligned}& 2\sum R_{2}R_{3}\ge\sum (2r_{1}+r_{2}+r_{3})R_{1}, \end{aligned}$$

(4.3)

$$\begin{aligned}& \sum R_{1}^{2}\ge\sum r_{1}(R_{2}+R_{3}). \end{aligned}$$

(4.4)

According to these two inequalities, by the same arguments used in the proof of Theorem 3.1, we deduce that the double inequality (4.2) holds.

Since the equalities of $w_{1}\ge r_{1}$, $w_{2}\ge r_{2}$, and $w_{3}\ge r_{3}$ are all valid if and only if P is the circumcenter of $\triangle ABC$, thus by the equality conditions of (2.1) and (3.1), it is easy to conclude that both equalities of (4.3) and (4.4) hold if and only if $\triangle ABC$ is equilateral and P is its center. Further, we know that the statement in Theorem 4.1 for the equalities is right. This completes the proof of Theorem 4.1. □

In particular, for $k=0$ in Theorem 4.1, we obtain the following.

Corollary 4.1

For any interior point P of $\triangle ABC$, we have

$$ \sum R_{1}\ge\frac{\sum(R_{2}+R_{3})(r_{1}+R_{1})}{\sum R_{1}}\ge2\sum r_{1}. $$

(4.5)

In a recent paper [33], Theorem 1, the author established the following sharpened version of the Erdös-Mordell inequality:

$$ \sum\frac{(r_{2}+r_{3})^{2}}{R_{1}}\le\sum R_{1}, $$

(4.6)

with equality holding if and only if $\triangle ABC$ is equilateral and P is its center or $\triangle ABC$ is a right isosceles triangle and P is its circumcenter.

Next, we apply inequality (4.6) to prove another refinement with one parameter for the Erdös-Mordell inequality.

Theorem 4.2

Let P be an interior point P of the triangle $ABC$ (P may lie on the boundary except the vertices of $ABC$) and let $k\ge1$ be a real number, then

$$ \sum R_{1}\ge\frac{1}{(k+1)^{2}}\sum \frac{(kR_{1}+r_{2}+r_{3})^{2}}{R_{1}}\ge 2\sum r_{1}. $$

(4.7)

Equalities in (4.7) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

We have

$$\begin{aligned} &\sum\frac{(kR_{1}+r_{2}+r_{3})^{2}}{R_{1}} \\ &\quad=\frac{k^{2}R_{1}^{2}+2kR_{1}(r_{2}+r_{3})+(r_{2}+r_{3})^{2}}{R_{1}} \\ &\quad=k^{2}\sum R_{1}+4k\sum r_{1}+ \sum\frac{(r_{2}+r_{3})^{2}}{R_{1}} \\ &\quad\le k^{2}\sum R_{1}+2k\sum R_{1}+\sum R_{1} \\ &\quad=(k+1)^{2}\sum R_{1}, \end{aligned}$$

where we used the Erdös-Mordell inequality (1.1) and inequality (4.6). Thus, the first inequality in (4.7) is proved (which in fact holds for $k\ge0$).

On the other hand, by $k\ge1$ and the Cauchy-Schwarz inequality we have

$$\begin{aligned} \sum\frac{(kR_{1}+r_{2}+r_{3})^{2}}{R_{1}}\ge\frac{ [\sum (kR_{1}+r_{2}+r_{3}) ]^{2}}{\sum R_{1}}=\frac{ (k\sum R_{1}+2\sum r_{1} )^{2}}{\sum R_{1}}. \end{aligned}$$

Thus, to prove the second inequality of (4.7), we only need to prove that

$$ \Bigl(k\sum R_{1}+2\sum r_{1} \Bigr)^{2}\ge2(k+1)^{2}\sum R_{1}\sum r_{1}. $$

(4.8)

By the hypothesis $k\ge1$ and the Erdös-Mordell inequality (1.1), we obtain

$$\begin{aligned} & \Bigl(k\sum R_{1}+2\sum r_{1} \Bigr)^{2}-2(k+1)^{2}\sum R_{1}\sum r_{1} \\ &\quad= \Bigl(\sum R_{1}-2\sum r_{1} \Bigr) \Bigl(k^{2}\sum R_{1}-2\sum r_{1} \Bigr) \\ &\quad\ge \Bigl(\sum R_{1}-2\sum r_{1} \Bigr)^{2}\ge0. \end{aligned}$$

Thus, inequality (4.8) and the second inequality in (4.7) are proved. According to the equality conditions of (1.1), (4.6) and the Cauchy-Schwarz inequality, we see that the equalities in (4.7) hold if and only if $\triangle ABC$ is equilateral and P is its center. This completes the proof of the Theorem 4.2. □

For $k=0$ in Theorem 4.2, we get the following.

Corollary 4.2

Let P be an interior point of the triangle $ABC$ (P may lie on the boundary except the vertices of $ABC$), then

$$ \sum R_{1}\ge\frac{1}{4}\sum \frac{(R_{1}+r_{2}+r_{3})^{2}}{R_{1}}\ge2\sum r_{1}. $$

(4.9)

In [28], for proving Oppenheim’s inequality (1.3), we presented the following inequality:

$$ R_{2}+R_{3}\ge2r_{1}+ \frac{(r_{2}+r_{3})^{2}}{R_{1}}, $$

(4.10)

with equality holds if and only if $b=c$ and P is its circumcenter of triangle $ABC$. We have also pointed out that the Erdös-Mordell inequality (1.1) can easily be obtained from (4.10) in [33]. Next, we shall use (4.10), the previous inequalities (2.7) and (4.4) to establish the following refinement of the Erdös-Mordell inequality.

Theorem 4.3

For any interior point P of the triangle $ABC$, the following inequalities hold:

$$ \sum R_{1}\ge\sqrt{\sum \bigl[R_{1}^{2}+2r_{1}R_{1}+(r_{2}+r_{3})^{2} \bigr]}\ge 2\sum r_{1}. $$

(4.11)

Equalities in (4.11) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

By inequality (4.10), we have

$$\sum R_{1}(R_{2}+R_{3})\ge2\sum R_{1}r_{1}+\sum(r_{2}+r_{3})^{2}, $$

so that

$$ 2\sum R_{2}R_{3}\ge2\sum r_{1}R_{1}+\sum(r_{2}+r_{3})^{2}. $$

(4.12)

Adding $\sum R_{1}^{2}$ to both sides of (4.12), we get

$$\Bigl(\sum R_{1} \Bigr)^{2}\ge\sum R_{1}^{2}+2\sum r_{1}R_{1}+\sum (r_{2}+r_{3})^{2}=\sum \bigl[R_{1}^{2}+2r_{1}R_{1}+(r_{2}+r_{3})^{2} \bigr], $$

which shows that the first inequality in (4.11) is true.

On the other hand, by inequalities (4.4), (2.7) and the Erdös-Mordell inequality we have

$$\begin{aligned} &\sum \bigl[R_{1}^{2}+2r_{1}R_{1}+(r_{2}+r_{3})^{2} \bigr] \\ &\quad=\sum R_{1}^{2}+2\sum r_{1}R_{1}+\sum(r_{2}+r_{3})^{2} \\ &\quad\ge\sum r_{1}(R_{2}+R_{3})+\sum r_{1}R_{1}+2\sum r_{2}r_{3}+ \sum(r_{2}+r_{3})^{2} \\ &\quad=\sum r_{1}\sum R_{1}+2\sum r_{2}r_{3}+\sum(r_{2}+r_{3})^{2} \\ &\quad\ge2 \Bigl(\sum r_{1} \Bigr)^{2}+2\sum r_{2}r_{3}+\sum(r_{2}+r_{3})^{2} \\ &\quad=4 \Bigl(\sum r_{1} \Bigr)^{2}. \end{aligned}$$

Thus, the second inequality in (4.11) holds.

In view of the equality conditions of (4.10), (2.7), and (1.1), we immediately conclude that the equalities in (4.11) hold if and only if $\triangle ABC$ is equilateral and P is its center. This completes the proof of Theorem 4.3. □

In the proof of the final theorem given in this section, we shall use the following well-known result (for proofs, see e.g., [5, 14, 19]).

Lemma 4.1

For any interior point P of the triangle $ABC$, we have

$$ \left \{ \textstyle\begin{array}{@{}l} aR_{1}\ge br_{3}+cr_{2}, \\ bR_{2}\ge cr_{1}+ar_{3}, \\ cR_{3}\ge ar_{2}+br_{1}. \end{array}\displaystyle \right . $$

(4.13)

Equalities in (4.13) successively hold if and only if P lies on the line AO, BO, and CO, where O is the circumcenter of the triangle $ABC$.

Theorem 4.4

For any interior point P of the triangle $ABC$, the following inequalities hold:

$$ \sum R_{1}\ge\sqrt{\sum(R_{2}+R_{3}) (r_{1}+R_{1})}\ge\sqrt{\frac{4}{3}\sum (r_{2}+R_{2}) (r_{3}+R_{3})}\ge2\sum r_{1}. $$

(4.14)

Equalities in (4.14) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

The first inequality in (4.14) is equivalent to the proved inequality (4.4), since

$$\Bigl(\sum R_{1} \Bigr)^{2}-\sum (R_{2}+R_{3}) (r_{1}+R_{1})=\sum R_{1}^{2}-\sum r_{1}(R_{2}+R_{3}). $$

By the previous inequality, (4.3), we have

$$\begin{aligned} &3\sum(R_{2}+R_{3}) (r_{1}+R_{1})-4 \sum(r_{2}+R_{2}) (r_{3}+R_{3})\\ &\quad=2\sum R_{2}R_{3}-\sum r_{1}(R_{2}+R_{3})-4\sum r_{2}r_{3}\\ &\quad\ge\sum(2r_{1}+r_{2}+r_{3})R_{1}- \sum r_{1}(R_{2}+R_{3})-4\sum r_{2}r_{3}\\ &\quad=\sum(r_{2}+r_{3})R_{1}+2\sum r_{1}R_{1}-\sum r_{1}(R_{2}+R_{3})-4 \sum r_{2}r_{3}\\ &\quad=2 \Bigl(\sum r_{1}R_{1}-2\sum r_{2}r_{3} \Bigr)\\ &\quad\ge0, \end{aligned}$$

where we used Lemma 2.2 in the last step. Thus, the second inequality in (4.14) is proved.

The third inequality in (4.14) is equivalent to

$$ \sum(r_{2}+R_{2}) (r_{3}+R_{3})\ge3 \Bigl(\sum r_{1} \Bigr)^{2}. $$

(4.15)

By Lemma 4.1 it is sufficient to prove that

$$\sum \biggl(r_{2}+\frac{cr_{1}+ar_{3}}{b} \biggr) \biggl(r_{3}+\frac {ar_{2}+br_{1}}{c} \biggr)\ge3 \Bigl(\sum r_{1} \Bigr)^{2}, $$

i.e.,

$$\sum a(cr_{1}+ar_{3}+br_{2}) (ar_{2}+br_{1}+cr_{3})-3abc \Bigl(\sum r_{1} \Bigr)^{2}\ge0, $$

which is equivalent to

$$ \sum a(b-c)^{2}r_{1}^{2}+ \sum \bigl(a^{3}+ab^{2}+ac^{2}+bc^{2}+b^{2}c-5abc \bigr)r_{2}r_{3}\ge0. $$

(4.16)

By the arithmetic-geometric means inequality, we have

$$a^{3}+ab^{2}+ac^{2}+bc^{2}+b^{2}c-5abc \ge0, $$

with equality holding if and only if $a=b=c$. Hence, inequalities (4.16) and then (4.15) are proved, while the equality in (4.15) occurs only when $\triangle ABC$ is equilateral and P is its center. This completes the proof of Theorem 4.4. □

5 Refinements of the Erdös-Mordell inequality II

The refinements of the Erdös-Mordell inequality, given in the last section, involve six segments $R_{1}$, $R_{2}$, $R_{3}$, $r_{1}$, $r_{2}$, $r_{3}$ but not the geometric elements of $\triangle ABC$. In this section, we use an unified method based on two lemmas (Lemmas 5.1 and 5.2 below) to establish three new refinements of the Erdös-Mordell inequality, which also cover the sides, the altitudes, and the medians of the triangle $ABC$ besides the above six segments.

The result of Theorem 4.1 prompts us to consider the following general refinements of the Erdös-Mordell inequality in the form:

$$ \sum R_{1}\ge\frac{2\sum k_{1}(r_{1}+R_{1})}{\sum k_{1}}\ge2\sum r_{1}, $$

(5.1)

where $k_{1}$, $k_{2}$, $k_{3}$ are positive real numbers.

Before offering the results of this kind of double inequalities, we first prove two related lemmas.

Lemma 5.1

If the positive real numbers $k_{1}$, $k_{2}$, $k_{3}$ form a triangle and satisfy

$$ \left \{ \textstyle\begin{array}{@{}l} k_{1}(b-c)^{2}+(k_{2}-k_{3})(b^{2}-c^{2})\ge0,\\ k_{2}(c-a)^{2}+(k_{3}-k_{1})(c^{2}-a^{2})\ge0,\\ k_{3}(a-b)^{2}+(k_{1}-k_{2})(a^{2}-b^{2})\ge0, \end{array}\displaystyle \right . $$

(5.2)

then for any interior point P of the triangle $ABC$ the following inequality holds:

$$ \sum R_{1}\ge\frac{2\sum k_{1}(r_{1}+R_{1})}{\sum k_{1}}. $$

(5.3)

Equality in (5.3) holds if and only if the equalities in (5.2) are all valid and P is the circumcenter of the triangle $ABC$.

Proof

Note that

$$\sum k_{1}\sum R_{1}-2\sum k_{1}(r_{1}+R_{1})=\sum(k_{2}+k_{3}-k_{1})R_{1}-2 \sum k_{1}r_{1}. $$

Inequality (5.3) is equivalent to

$$ \sum(k_{2}+k_{3}-k_{1})R_{1} \ge2\sum k_{1}r_{1}, $$

(5.4)

which is obviously a generalized form of the Erdös-Mordell inequality. Since $k_{1}$, $k_{2}$, $k_{3}$ form a triangle, we have $k_{2}+k_{3}-k_{1}>0$ etc. Thus, by Lemma 4.1, to prove (5.4) we only need to prove that

$$\sum(k_{2}+k_{3}-k_{1}) \frac{br_{3}+cr_{2}}{a}-2\sum k_{1}r_{1}\ge0, $$

i.e.,

$$\sum bc(k_{2}+k_{3}-k_{1}) (br_{3}+cr_{2})-2abc\sum k_{1}r_{1} \ge0. $$

Expending and rearranging gives the equivalent inequality

$$ \sum a \bigl[k_{1}(b-c)^{2}+(k_{2}-k_{3}) \bigl(b^{2}-c^{2}\bigr) \bigr]r_{1}\ge0, $$

(5.5)

which is clearly true if (5.2) is valid. From Lemma 4.1 and (4.16), one sees that the equality in (4.15) occurs if and only if the equalities in (5.2) are all valid and P is the circumcenter of the triangle $ABC$. This completes the proof of Lemma 5.1. □

Lemma 5.2

If the positive real numbers $k_{1}$, $k_{2}$, $k_{3}$ satisfy

$$ \left \{ \textstyle\begin{array}{@{}l} (b-c)(bk_{3}-ck_{2})\ge0,\\ (c-a)(ck_{1}-ak_{3})\ge0,\\ (a-b)(ak_{2}-bk_{1})\ge0, \end{array}\displaystyle \right . $$

(5.6)

then for any interior point P of the triangle $ABC$ the following inequality holds:

$$ \frac{\sum k_{1}(r_{1}+R_{1})}{\sum k_{1}}\ge\sum r_{1}. $$

(5.7)

Equality in (5.7) holds if and only if equalities in (5.6) are all valid and P is the circumcenter of the triangle $ABC$.

Proof

It is easy to know that inequality (5.7) is equivalent to

$$ \sum k_{1}R_{1}-\sum (k_{2}+k_{3})r_{1}\ge0, $$

(5.8)

which is obviously a generalized form of the Erdös-Mordell inequality. Since $k_{1},k_{2},k_{3}>0$, by Lemma 4.1, to prove the above inequality we need to prove that

$$\sum k_{1}\frac{br_{3}+cr_{2}}{a}-\sum(k_{2}+k_{3})r_{1} \ge0. $$

Multiplying both sides by $abc$ and rearranging gives the equivalent inequality

$$ \sum a(b-c) (bk_{3}-ck_{2})r_{1} \ge0, $$

(5.9)

which is clearly true if (5.6) is valid. By Lemma 4.1 and (5.6), we conclude that the equality conditions mentioned in the lemma is true. This completes the proof of Lemma 5.2. □

Lemma 5.1 and Lemma 5.2 show that if the positive real numbers $k_{1}$, $k_{2}$, $k_{3}$ form a triangle and satisfy (5.2) and (5.6), then the refinement (5.1) of the Erdös-Mordell inequality holds. Next, we shall establish three refinement of the Erdös-Mordell inequality by applying this conclusion.

Theorem 5.1

For any interior point P of the triangle $ABC$ and arbitrary non-negative real numbers λ and μ (λ, μ not both zero) the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(\lambda a+\mu b+\mu c)(r_{1}+R_{1})}{(\lambda +2\mu)\sum a}\ge2\sum r_{1}. $$

(5.10)

If $\lambda=0$, then the first equality in (5.10) holds if and only if P is the circumcenter of $ABC$; If $\mu=0$, then the second equality in (5.10) holds if and only if P is the circumcenter of $ABC$; in other cases, the equalities hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

We first prove the first inequality in (5.10) by means of Lemma 5.1. Let $\lambda a+\mu b+\mu c=k_{1}$, $\lambda b+\mu c+\mu a=k_{2}$, and $\lambda c+\mu a+\mu b=k_{3}$, where $\lambda \ge0$, $\mu\ge0$ and λ, μ not both zero. Then

$$k_{2}+k_{3}-k_{1}=\lambda(b+c-a)+2\mu a>0. $$

Similarly, we have $k_{3}+k_{1}-k_{2}>0$ and $k_{1}+k_{2}-k_{3}>0$. Thus, $k_{1}$, $k_{2}$, $k_{3}$ form a triangle, and $\sum k_{1}=(\lambda+2\mu)\sum a$. A short calculation gives

$$\begin{aligned} k_{1}(b-c)^{2}+(k_{2}-k_{3}) \bigl(b^{2}-c^{2}\bigr)=\lambda(a+b+c) (b-c)^{2} \ge0, \end{aligned}$$

and two similar relations hold. We hence by Lemma 5.1 conclude that the first inequality in (5.10) holds.

We now use Lemma 5.2 to prove the second inequality in (5.10). Under the above assumptions, we easily get

$$\begin{aligned} (b-c) (bk_{3}-ck_{2})=\mu(a+b+c) (b-c)^{2}\ge0. \end{aligned}$$

Similarly, we have $(c-a)(ck_{1}-ak_{3})\ge0$ and $(a-b)(ak_{2}-bk_{1})\ge0$. Thus, by Lemma 5.2, the second inequality in (5.10) is proved.

It is easy to see that the first inequality in (5.10) when $\lambda =0$ and the second inequality in (5.10) when $\mu=0$ are both equivalent to

$$ \sum aR_{1}\ge\sum(b+c)r_{1}, $$

(5.11)

which follows from (4.13). By Lemma 4.1, we see that the equalities (5.10) in the above two cases holds if and only if P is the circumcenter of $\triangle ABC$. For other cases, by Lemma 5.1 and Lemma 5.2 we easily conclude that the equalities in (5.10) are valid if and only if $a=b=c$ and P is the circumcenter of $\triangle ABC$, i.e., $\triangle ABC$ is equilateral and P is its center. This completes the proof of Theorem 5.1. □

In (5.10), for $\mu=0$, we obtain the following.

Corollary 5.1

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}\ge \frac{2\sum a(r_{1}+R_{1})}{\sum a}\ge2\sum r_{1}. $$

(5.12)

Remark 5.1

By applying Lemma 5.1 and Lemma 5.2, it is easy to prove the following exponential generalization of (5.12):

$$ \sum R_{1}\ge\frac{2\sum a^{k}(r_{1}+R_{1})}{\sum a^{k}}\ge2\sum r_{1}, $$

(5.13)

where k is a real number such that $0< k\le1$.

In (5.10), for $\lambda=0$, we obtain

Corollary 5.2

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}\ge \frac{\sum(b+c)(r_{1}+R_{1})}{\sum a}\ge2\sum r_{1}. $$

(5.14)

Since $r_{1}+R_{1}\ge h_{a}=r\sum a/a$, where r is the inradius of $\triangle ABC$. Thus, by the first inequality of (5.12), we obtain a lower bound of $\sum R_{1}$, namely

Corollary 5.3

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}\ge r\sum \frac{b+c}{a}. $$

(5.15)

Motivated by Theorem 5.1, we find that if the lengths of the sides a, b, c are replaced by the corresponding altitudes $h_{a}$, $h_{b}$, $h_{c}$ of $\triangle ABC$ in (5.10), respectively, then the inequalities hold for $\mu>0$ and $2\mu\ge\lambda\ge0$, i.e., the following double inequality holds:

$$ \sum R_{1}\ge \frac{2\sum(\lambda h_{a}+\mu h_{b}+\mu h_{c})(r_{1}+R_{1})}{(\lambda+2\mu)\sum h_{a}}\ge2\sum r_{1}, $$

(5.16)

which is required to prove. If we put $\frac{\lambda}{\mu}=k$, then (5.16) becomes the double inequality (5.17) below.

Theorem 5.2

Let k be a real number such that $0\le k\le2$, then for any interior point P of the triangle $ABC$ the following inequalities hold:

$$ \sum R_{1}\ge \frac{2\sum(kh_{a}+h_{b}+h_{c})(r_{1}+R_{1})}{(k+2)\sum h_{a}}\ge 2\sum r_{1}. $$

(5.17)

Equalities in (5.17) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

We put

$$kh_{a}+h_{b}+h_{c}=k_{1},\qquad kh_{b}+h_{c}+h_{a}=k_{2},\qquad kh_{c}+h_{a}+h_{b}=k_{3}. $$

Then by the given condition $0\le k\le2$ we have

$$k_{2}+k_{3}-k_{1}=(2-k)h_{a}+k(h_{b}+h_{c})>0. $$

Similarly, we get $k_{3}+k_{1}-k_{2}>0$ and $k_{1}+k_{2}-k_{3}>0$. Thus, $k_{1}$, $k_{2}$, $k_{3}$ form a triangle and $\sum k_{1}=(k+2)\sum h_{a}$. Also, it is easy to check the following identity:

$$\begin{aligned} &k_{1}(b-c)^{2}+(k_{2}-k_{3}) \bigl(b^{2}-c^{2}\bigr)\\ &\quad=2(b-c) (bh_{c}-ch_{b})+k(b-c) (bh_{a}+bh_{b}+ch_{b}-bh_{c}-ch_{a}-ch_{c}). \end{aligned}$$

Thus, by $2\ge k\ge0$, $(b-c)(bh_{c}-ch_{b})\ge0$, and $bh_{b}=ch_{c}$, we have

$$\begin{aligned} &k_{1}(b-c)^{2}+(k_{2}-k_{3}) \bigl(b^{2}-c^{2}\bigr)\\ &\quad\ge k(b-c) (bh_{c}-ch_{b})+k(b-c) (bh_{a}+ch_{b}-bh_{c}-ch_{a})\\ &\quad=kh_{a}(b-c)^{2}\ge0. \end{aligned}$$

Two inequalities similar to $k_{1}(b-c)^{2}+(k_{2}-k_{3})(b^{2}-c^{2})\ge0$ of course hold. Thus, the first inequality in (5.17) is proved by Lemma 5.1.

On the other hand, it is easy to get the following identity:

$$\begin{aligned} (b-c) (bk_{3}-ck_{2})=(h_{a}+kh_{b}+kh_{c}) (b-c)^{2}. \end{aligned}$$

Hence, $(b-c)(bk_{3}-ck_{2})\ge0$ is true and its two analogs hold for $k\ge 0$. Thus, by Lemma 5.2, the second inequality in (5.17) is valid for $k\ge0$.

Summarizing, we deduce that the double inequality (5.17) holds for $0\le k\le2$. In addition, by Lemma 5.1 and Lemma 5.2, we easily conclude that the equalities in (5.17) occur hold only when $\triangle ABC$ is equilateral and P is its center. The proof of Theorem 5.2 is completed. □

In particular, for $k=0$ in Theorem 5.2, we obtain

Corollary 5.4

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}\ge \frac{\sum(h_{b}+h_{c})(r_{1}+R_{1})}{\sum h_{a}}\ge2\sum r_{1}. $$

(5.18)

Finally, we give a dual result of Theorem 5.2, which shows that if we replace the altitudes $h_{a}$, $h_{b}$, $h_{c}$ by the corresponding medians $m_{a}$, $m_{b}$, $m_{c}$ in (5.17) then the inequalities still hold, i.e., we have the following conclusion:

Theorem 5.3

Let k be a real number such that $0\le k\le2$, then for any interior point P of $\triangle ABC$ the following inequalities hold:

$$ \sum R_{1}\ge \frac{2\sum(km_{a}+m_{b}+m_{c})(r_{1}+R_{1})}{(k+2)\sum m_{a}}\ge 2\sum r_{1}. $$

(5.19)

Equalities in (5.19) hold if and only if $\triangle ABC$ is equilateral and P is its center.

Proof

In order to prove double inequality (5.19), we first prove the following two inequalities involving the medians and the sides of the triangle, i.e.,

$$ (b-c) (bm_{c}-cm_{b})\ge0, $$

(5.20)

and

$$ (b-c) \bigl[b(m_{a}+m_{b})-c(m_{c}+m_{a}) \bigr]\ge0. $$

(5.21)

Inequality (5.20) can be obtained directly from the following identity:

$$ 2(b-c) (bm_{c}-cm_{b}) (bm_{c}+cm_{b})=(b+c) \bigl(a^{2}+b^{2}+c^{2}\bigr) (b-c)^{2}, $$

(5.22)

which is easily checked by using the known formulas $4m_{b}^{2}=2(c^{2}+a^{2})-b^{2}$ and $4m_{c}^{2}=2(a^{2}+b^{2})-c^{2}$. For inequality (5.21), expanding and rearranging gives its equivalent form:

$$ m_{a}\bigl(b^{2}+c^{2} \bigr)+m_{b}b^{2}+m_{c}c^{2}-bc(2m_{a}+m_{b}+m_{c}) \ge0, $$

(5.23)

which is needed to prove. According to the median-dual transformation (cf. [4], pp.109-111), we know that the above inequality is equivalent to

$$ a\bigl(m_{b}^{2}+m_{c}^{2} \bigr)+bm_{b}^{2}+cm_{c}^{2}-m_{b}m_{c}(2a+b+c) \ge0. $$

(5.24)

Using the formulas for the medians $m_{b}$ and $m_{c}$ again, it is easy to verify the following identity:

$$\begin{aligned} & \bigl[a\bigl(m_{b}^{2}+m_{c}^{2} \bigr)+bm_{b}^{2}+cm_{c}^{2} \bigr]^{2}-(2a+b+c)^{2}(m_{b}m_{c})^{2} \\ &\quad=\frac{3}{16}(b+c) (b+c-a) (b-c)^{2}\bigl(a^{2}+b^{2}+c^{2}+3ab+3ac \bigr), \end{aligned}$$

(5.25)

which shows that (5.24) holds true. Therefore, inequalities (5.23) and (5.21) are proved. Also, we easily known that both equalities of (5.20) and (5.21) occur if and only if $b=c$.

We now apply Lemma 5.1 and Lemma 5.2 to prove the double inequality (5.19). Let

$$km_{a}+(m_{b}+m_{c})=k_{1},\qquad km_{b}+(m_{c}+m_{a})=k_{2},\qquad km_{c}+(m_{a}+m_{b})=k_{3}, $$

then it is not difficult to show that $k_{1}$, $k_{2}$, $k_{3}$ can be form a triangle if $0\le k\le2$ and we have $\sum k_{1}=(k+2)\sum m_{a}$. By Lemma 5.1, to prove the first inequality in (5.19), it remains to show that $k_{1}(b-c)^{2}+(k_{2}-k_{3})(b^{2}-c^{2})\ge0$. Taking into account that $2\ge k\ge0$, by (5.20) and (5.21) we have

$$\begin{aligned} &k_{1}(b-c)^{2}+(k_{2}-k_{3}) \bigl(b^{2}-c^{2}\bigr) \\ &\quad=2(b-c) (bm_{c}-cm_{b})+k(b-c) (bm_{a}+bm_{b}+cm_{b}-bm_{c}-cm_{a}-cm_{c}) \\ &\quad\ge k(b-c) (bm_{c}-cm_{b})+k(b-c) (bm_{a}+bm_{b}+cm_{b}-bm_{c}-cm_{a}-cm_{c}) \\ &\quad=k(b-c) \bigl[b(m_{a}+m_{b})-c(m_{c}+m_{a}) \bigr]\ge0, \end{aligned}$$

as claimed. Hence, the first inequality in (5.19) is proved.

The second inequality in (5.19) can also easily be proved by Lemma 5.2. With the above assumptions, it is easy to obtain the following identity:

$$\begin{aligned} &(b-c) (bk_{3}-ck_{2})\\ &\quad=(b-c) \bigl[b(m_{a}+m_{b})-c(m_{a}+m_{c}) \bigr]+k(b-c) (bm_{c}-cm_{b}), \end{aligned}$$

which together with (5.20) and (5.21) shows that $(b-c)(bk_{3}-ck_{2})\ge0$ holds for $k\ge0$. Thus, we finish the proof of the second inequality in (5.19).

The equality conditions of (5.19) follow easily from Lemma 5.1 and 5.2. This completes the proof of Theorem 5.3. □

For $k=0$ in (5.19), we get the following counterpart of (5.18).

Corollary 5.5

For any interior point P of the triangle $ABC$, we have

$$ \sum R_{1}\ge \frac{\sum(m_{b}+m_{c})(r_{1}+R_{1})}{\sum m_{a}}\ge2\sum r_{1}. $$

(5.26)

6 Open problems

For the inequalities established in this paper, we can propose a lot of new problems. We next introduce some related conjectures as open problems, which have been checked by computer.

The first inequalities of (1.4) and (2.8) prompt the author to propose the following conjecture.

Conjecture 6.1

For any interior point P of $\triangle ABC$, we have

$$ \sum R_{1}(2r_{1}+r_{2}+r_{3}) \ge2\sum h_{a}r_{1}+2\sum r_{2}r_{3}. $$

(6.1)

If inequality (6.1) holds true, then we have the following refinement of the Oppenheim inequality (1.3):

$$\begin{aligned} \sum R_{2}R_{3}&\ge \frac{1}{2}\sum R_{1}(2r_{1}+r_{2}+r_{3}) \ge\sum h_{a}r_{1}+\sum r_{2}r_{3} \\ & \ge\sum(r_{3}+r_{1}) (r_{1}+r_{2}). \end{aligned}$$

(6.2)

in which the last inequality is proved by the author in [28].

From the refinement (4.1) of the Erös-Mordell inequality, we see that the following inequality holds:

$$ \sum\sqrt{a^{2}+4r_{1}^{2}} \ge4\sqrt{\sum h_{a}r_{1}}, $$

(6.3)

which is difficult to prove directly. Here, we conjecture that the above inequality can be strengthened as follows.

Conjecture 6.2

For any interior point P of $\triangle ABC$, we have

$$ \sum\sqrt{a^{2}+4r_{1}^{2}} \ge4\sqrt{\sum m_{a}r_{1}}. $$

(6.4)

The following conjecture is similar to Theorem 4.1.

Conjecture 6.3

Let k be a real number such that $0.48\le k\le1.36$, then for any interior point P of $\triangle ABC$ the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(kr_{1}+r_{2}+r_{3})(R_{1}+r_{1})}{(k+2)\sum r_{1}}\ge 2\sum r_{1}. $$

(6.5)

Another conjecture with one parameter, which comes from considering generalizations of the previous double inequality (5.12), is as follows.

Conjecture 6.4

Let k be a real number such that $0< k<4$, then for any interior point P of $\triangle ABC$ the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(R_{1}+r_{1})\sqrt{a^{2}+kr_{1}^{2}}}{\sum \sqrt{a^{2}+kr_{1}^{2}}}\ge2\sum r_{1}. $$

(6.6)

Many years ago, the author found the following interesting ‘r-w’ phenomenon: If an inequality involving the segments $r_{1}$, $r_{2}$, $r_{3}$ and other geometric elements holds for any interior P of $\triangle ABC$, then the inequality from replacing $r_{1}$, $r_{2}$, $r_{3}$ by $w_{1}$, $w_{2}$, $w_{3}$ in the original inequality, respectively, often still holds for either any or acute triangle $ABC$. Based on this phenomenon and with the verification by computer, we can propose some dual inequalities for the results presented in this paper. Here, we only give one example, which is a dual conjecture of Theorem 5.1 as follows.

Conjecture 6.5

For any interior point P of $\triangle ABC$ and non-negative real numbers λ and μ (λ, μ not both zero), the following inequalities hold:

$$ \sum R_{1}\ge\frac{2\sum(\lambda a+\mu b+\mu c)(w_{1}+R_{1})}{(\lambda +2\mu)\sum a}\ge2\sum w_{1}. $$

(6.7)

When $\lambda=0$, the first inequality in (6.7) actually is equivalent to the following beautiful inequality:

$$ \sum aR_{1}\ge\sum(b+c)w_{1}, $$

(6.8)

which is clearly sharper than the previous inequality (5.11) but has not been proved at present.