To prove our main results, we need the following lemmas.

### Lemma 1

*Let*
*K*, *W*
*be two convex bodies in*
\(\mathbb{R}^{2}\). *Let*
\(p_{K}(\theta)\)
*and*
\(p_{W}(\theta)\)
*be support functions of*
*K*
*and*
*W*, *respectively*. *If*
\(W\subseteq K\)
*such that*
\(p_{W}(\theta)\ne p_{K}(\theta)\), \(\theta\in[\theta_{0},\theta_{0}+\pi]\)
*for some*
\(\theta_{0}\), *then there exist*
\(\epsilon >0\)
*and*
\(v\in S^{1}\)
*such that*
\(W+\epsilon\cdot v\subset K\).

### Proof

Choose vectors \(u, v\in S^{1}\) corresponding to angles \(\theta, \theta_{0}+\frac{\pi}{2}\), respectively. If \(\theta\in [\theta_{0}, \theta_{0}+\pi]^{c}\), the complement of \([\theta_{0}, \theta_{0}+\pi]\), since the angle between *u* and *v* is strictly greater than \(\frac{\pi}{2}\), hence \(u\cdot v < 0\). By (17), we have

$$\begin{aligned} p_{W+\epsilon\cdot v}(u)=p_{W}(u)+\epsilon(u\cdot v)< p_{K}(u) \end{aligned}$$

(23)

for \(\epsilon>0\).

If \(\theta\in[\theta_{0},\theta_{0}+\pi]\), since \(p_{W}(\theta)\ne p_{K}(\theta)\) when \(\theta\in[\theta_{0},\theta_{0}+\pi]\) and \(W \subseteq K\), we have

For \(\epsilon>0\) small enough, we have

$$\begin{aligned} p_{W+\epsilon\cdot v}(u)=p_{W}(u)+\epsilon(u\cdot v)< p_{K}(u). \end{aligned}$$

(24)

Hence

$$p_{W+\epsilon\cdot v}(u)< p_{K}(u), $$

for all \(u\in S^{1}\), that is, \(W+\epsilon\cdot v\subset K\). □

### Lemma 2

*Let*
\(K, W\)
*be two oval bodies in*
\(\mathbb{R}^{2}\). *Let*
\(r_{W}\), \(R_{W}\)
*be*, *respectively*, *the W*-*inradius and W*-*outradius of*
*K*. *Then the equation*
\(A_{K,W}(t)=0\)
*has two roots*
\(t_{1}, t_{2}\)
*such that*

$$\begin{aligned} t_{1}\leq-R_{W}\leq-r_{W} \leq t_{2}< 0. \end{aligned}$$

(25)

*Each inequality in* (25) *holds as an equality if and only if*
*K*
*and*
*W*
*are homothetic*. *In particular*, *when*
\(r_{W}\leq t\leq R_{W}\),

$$\begin{aligned} A_{K,W}(-t)\leq0. \end{aligned}$$

(26)

*Inequality* (26) *is strict whenever*
\(r_{W}< t< R_{W}\). *When*
\(t=r_{W}\)
*or*
\(t=R_{W}\), *equality will occur in* (26) *if and only if*
*K*
*and*
*W*
*are homothetic*.

### Proof

There is at least one point where \(\partial(r_{W}W)\) is tangent to *∂K* for \(\theta\in[\theta_{0}, \theta_{0}+\pi]\) with all \(\theta_{0}\). If the conclusion fails, that is, there exists \(\theta_{0}\) such that \(p_{r_{W}W}(\theta) \ne p_{K}(\theta) \) for \(\theta\in[\theta_{0}, \theta_{0}+\pi]\), choose the vector *v* corresponding to the angle \(\theta_{0}+\frac{\pi}{2}\). By Lemma 1, if we move \(r_{W}W\) by *v* for \(\epsilon>0\) small enough, then \(r_{W}W+\epsilon\cdot v\) continues to lie in the interior of *K* and has no points of tangency. This contradicts the maximality of \(r_{W}\).

By integration by parts we have

$$\begin{aligned} A_{K,W}(-r_{W})&=\frac{1}{2} \int_{S^{1}}(p_{K}-r_{W}{p_{W}} ) \bigl(p_{K}+p_{K}^{\prime\prime}-r_{W} \bigl({p_{W}} +p_{W} ^{\prime\prime}\bigr)\bigr)\,d\theta \\ &=\frac{1}{2} \int_{S^{1}}(p_{K}-r_{W}{p_{W}} )^{2}\,d\theta+\frac{1}{2} \int_{S^{1}}(p_{K}-r_{W}{p_{W}} ) \bigl(p_{K}^{\prime\prime}-r_{W}p_{W} ^{\prime\prime}\bigr)\,d\theta \\ &= \frac{1}{2} \int_{S^{1}}(p_{K}-r_{W}{p_{W}} )^{2}\,d\theta-\frac{1}{2} \int_{S^{1}}\bigl(p_{K}^{\prime}-r_{W}p_{W} ^{\prime}\bigr)^{2}\,d\theta. \end{aligned}$$

(27)

Let \(\theta_{1}, \theta_{2},\ldots,\theta_{N}\) be points where \(\partial(r_{W}W)\) are tangent to *∂K*. We can break up the right-hand side of (27) into integrals over the intervals [\(\theta_{i},\theta_{i+1}\)] (\(1\leq i\leq N-1\)). Since every set \([\theta, \theta+\pi]\) contains a point where \(r_{W}W\) is tangent to *K*, we have

$$\theta_{i+1}-\theta_{i}\leq\pi. $$

Let

$$f=p_{K}-r_{W}{p_{W}}, $$

then

at each point of tangency. Applying inequality (22) in the Poincaré lemma, we have

$$\begin{aligned} A_{K,W}(-r_{W})\leq0, \end{aligned}$$

(28)

where the equality holds if and only if

$$p_{K}-r_{W} {p_{W}}=c\sin\biggl( \frac{\pi\theta}{\alpha}\biggr). $$

Since the convex body *K* contains *tW*, then

$$c\sin\biggl(\frac{\pi\theta}{\alpha}\biggr)=p_{K}-r_{W}{p_{W}} \geq0 $$

for all *θ*. This leads to

that is, *K* and *W* are homothetic. In a similar way, we have

$$\begin{aligned} A_{K,W}(-R_{W})\leq0, \end{aligned}$$

(29)

where the equality holds if and only if *K* and *W* are homothetic. Thus the equation \(A_{K,W}(t)=0\) has two roots \(t_{1}, t_{2}\), and

$$t_{1}+t_{2}=-\frac{ L_{K,W} }{A_{W}}< 0, \qquad t_{1}t_{2}= \frac{A_{K}}{A_{W}}>0, $$

and therefore

$$t_{1}< 0,\qquad t_{2}< 0. $$

Therefore

$$t_{1}\leq-R_{W}\leq-r_{W}\leq t_{2}. $$

In particular, according to (5), when \(r_{W}\leq t\leq R_{W}\), we have

If \(r_{W} < t< R_{W}\), then \(t_{1}<-t<t_{2}\). Inequality (26) is strict. Therefore, equality occurs in (26) only when \(t=r_{W}\) or \(t=R_{W}\), that is, *K* and *W* are homothetic. Lemma 2 is proved. □

### Remark

Inequality (26) has been mentioned in Green and Osher’s work (cf. [1]) without proof. For general convex bodies, Luo, Xu and Zhou [17] have also obtained inequality (26) by the integral geometry method. However, it is difficult to obtain the equality condition of inequality (26) for general convex bodies. Via the method of convex geometric analysis, a complete proof of inequality (26) with equality condition is given in [9].

By (28) or (29), the sufficient condition for root existence of equation \(A_{K, W}(t)=0\) is that the discriminant of \(A_{K, W}(t)=0\) is non-negative. We obtain the following Wulff isoperimetric inequality.

### Corollary 1

*Let*
\(K, W\)
*be two oval bodies in*
\(\mathbb{R}^{2}\)
*with areas*
\(A_{K}, A_{W}\), *then*

$$L_{K,W}^{2}-4A_{K}A_{W}\geq0, $$

*the equality holds if and only if*
*K*
*and*
*W*
*are homothetic*.

### Proof of Theorem 1

By inequalities (28), (29), we have, respectively,

$$\begin{aligned} &A_{K}- L_{K,W} r_{W}+A_{W}r_{W}^{2} \leq0 , \end{aligned}$$

(30)

$$\begin{aligned} &A_{K}- L_{K,W} R_{W}+A_{W}R_{W}^{2} \leq0 . \end{aligned}$$

(31)

Then inequalities (30), (31) can be, respectively, rewritten as

$$\begin{aligned} &{-}2A_{K}A_{W}\geq2A_{W}^{2}r_{W}^{2}-2 L_{K,W} r_{W}A_{W}, \\ &{-}2A_{K}A_{W}\geq2A_{W}^{2}R_{W}^{2}-2 L_{K,W} R_{W}A_{W}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} L_{K,W} ^{2}-4A_{K}A_{W}\geq{}& L_{K,W} ^{2}+2A_{W}^{2} r_{W}^{2}+2A_{W}^{2}R_{W}^{2}-2 L_{K,W} r_{W}A_{W}-2 L_{K,W} R_{W}A_{W} \\ ={}&A_{W}^{2}r_{W}^{2}+A_{W}^{2}R_{W}^{2}-2A_{W}^{2}r_{W}R_{W}+ L_{K,W} ^{2}+A_{W}^{2}r_{W}^{2}+A_{W}^{2}R_{W}^{2} \\ &{}+2A_{W}^{2}r_{W}R_{W}-2 L_{K,W} r_{W}A_{W}-2 L_{K,W} R_{W}A_{W} \\ ={}&A_{W}^{2}(R_{W}-r_{W})^{2}+(A_{W}R_{W}+A_{W}r_{W}- L_{K,W} )^{2}, \end{aligned}$$

where the equality holds if and only if the equalities of (28), (29) hold, that is, *K* and *W* are homothetic. This proves inequality (11).

Inequalities (30), (31) can also be rewritten, respectively, as follows:

$$\begin{aligned} &L_{K,W} r_{W}\geq A_{W}r_{W}^{2}+A_{K}, \\ &L_{K,W} R_{W}\geq A_{W}R_{W}^{2}+A_{K}. \end{aligned}$$

Therefore

$$\begin{aligned} L_{K,W} ^{2}r_{W}R_{W}-4A_{K}A_{W}r_{W}R_{W} \geq{}& A_{W}^{2}r_{W}^{2}R_{W}^{2}+A_{K}A_{W}r_{W}^{2}+A_{K}A_{W}R_{W}^{2}+A_{K}^{2} \\ &{}-4A_{K}A_{W}r_{W}R_{W} \\ ={}&A_{K}A_{W}(R_{W}-r_{W})^{2}+(R_{W}r_{W}A_{W}-A_{K})^{2}. \end{aligned}$$

Hence, we have

$$L_{K,W} ^{2}-4A_{K}A_{W}\geq \frac{A_{K}A_{W}(R_{W}-r_{W})^{2}}{R_{W}r_{W}}+\frac{(R_{W}r_{W}A_{W}-A_{K})^{2}}{R_{W}r_{W}}, $$

where the equality holds if and only if *K* and *W* are homothetic. Inequality (12) is proved. □

Let *W* be the unit disc, then \(L_{K,W}^{2}=L_{K}^{2}\), \(A_{W}=\pi\). Therefore we have the following.

### Corollary 2

*Let*
*K*
*be an oval body in*
\(\mathbb{R}^{2}\)
*with area*
\(A_{K}\)
*and perimeter*
\(L_{K}\). *Let*
*r*
*and*
*R*
*be*, *respectively*, *the radius of the maximum inscribed disc and the radius of the minimum circumscribed disc of*
*K*. *Then*

$$\begin{aligned} &L_{K}^{2}-4\pi A_{K}\geq \pi^{2}(R-r)^{2}+( \pi R+\pi r- L_{K} )^{2}, \end{aligned}$$

(32)

$$\begin{aligned} &L_{K}^{2}-4\pi A_{K}\geq \frac{\pi A_{K}(R-r)^{2}}{Rr}+ \frac{(\pi Rr-A_{K})^{2}}{Rr}. \end{aligned}$$

(33)

*Each equality holds if and only if*
*K*
*is a disc*.

It should be noted that (32) is obtained in [24], which is stronger than the Bonnesen isoperimetric inequality (4).