Application 4.1 Let () such that
If , then
with equality for , and also for and (or any cyclic permutation).
Proof
Rewrite the desired inequality as
where
We have
where
There are two cases to consider.
Case 1: . For , , we have
since
Therefore, f is convex for , . By HCF-Theorem, we only need to show that for all which satisfy . According to Remark 2.4, this is true if for and , where
Indeed, we have
and
Case 2: . Since
and
from the expression of it follows that f is decreasing on and increasing on , where
On the other hand, for , we have
since
Thus, f is convex on . By PCF-Theorem, we only need to show that for all such that . We have proved this before (at Case 1). □
Application 4.2 If () are real numbers such that
then [6]
with equality for , and also for and (or any cyclic permutation).
Proof The desired inequality is true for since
and
Consider further that and rewrite the desired inequality as
where
We have
and
where
From the expression of , it follows that f is decreasing on and increasing on , where
On the other hand, for , we have
and hence . Since , f is convex on . By PCF-Theorem, we only need to show that for all which satisfy . According to Remark 2.4, this is true if for and , where
Indeed, we have
and
□
Application 4.3 Let () be positive real numbers such that
If , then [6]
with equality for .
Proof
Using the substitutions
and
the desired inequality becomes
where and . Clearly, if this inequality holds for , then it holds for any . Therefore, we need only to consider the case , when , and the desired inequality is equivalent to
There are two cases to consider: and .
Case 1: . By Bernoulli’s inequality, we have
and hence
Consequently, it suffices to prove that
For , this inequality is an equality. Otherwise, for , we rewrite the inequality as
which follows from the AM-HM inequality as follows:
Case 2: . Write the desired inequality as
where
We have
and
where
From the expression of , it follows that f is decreasing on and increasing on , where
On the other hand, for , we have
and hence
Thus, , and hence f is convex on . By PCF-Theorem and Remark 2.3, we need to show that for all positive x, y which satisfy and . Consider the nontrivial case where and write the inequality as follows:
Since and , it suffices to show that
which is equivalent to
where
By the weighted AM-GM inequality, we have
and hence h is strictly increasing. Since , we get
Thus, it suffices to show that
Putting , , we write this inequality as
By Bernoulli’s inequality,
So, we only need to show that , which is clearly true. □
Application 4.4 Let () be positive real numbers such that
If , then [6]
with equality for .
Proof
According to the power mean inequality, we have
for all . Thus, it suffices to prove the desired inequality for
Rewrite the desired inequality as
where
We have
From the expression of , it follows that f is decreasing on and increasing on , where
In addition, we claim that f is convex on . Indeed, since and
we have for . Therefore, by PCF-Theorem and Remark 2.3, we only need to show that
for and . We have
Also, from
we get
Therefore, by Bernoulli’s inequality, we have
□
Application 4.5 If a, b, c are positive real numbers such that , then
with equality for and also for (or any cyclic permutation).
Proof
Write the desired inequality as
where and
where , . From
it follows that g is decreasing on and increasing on , where
We have
where
We will show that for , and hence f is convex for
We have
Since
is increasing,
h is increasing, and hence
By PCF-Corollary, we only need to prove that for ; that is,
where
Indeed, we have
□
Application 4.6 If a, b, c are positive real numbers such that , then [6]
with equality for and also for (or any cyclic permutation).
Proof
Write the desired inequality as
where and
From
it follows that g is decreasing on and increasing on , where
We have
where
We will show that for , and hence f is convex for
We have
By PCF-Corollary, we only need to prove that for ; that is,
Since the last inequality is true, the proof is completed. □
Application 4.7 Let a, b, c be positive real numbers such that . If
then [6]
with equality for . If , then the equality holds also for and (or any cyclic permutation). If , then the equality holds also for and (or any cyclic permutation).
Proof
The desired inequality is equivalent to
Thus, it suffices to prove this inequality for only . On the other hand, replacing a, b, c by , , , the inequality becomes
Thus, we only need to prove the desired inequality for . Write this inequality as
where and
From
it follows that g is decreasing on and increasing on , where
We have
where
We will show that for , and hence f is convex for
Indeed, since
we have
By PCF-Corollary, we only need to prove that for ; that is,
The last inequality is clearly true, and the proof is completed. □
Application 4.8 If a, b, c are positive real numbers and , then [6]
with equality for . If , then the equality holds also for (or any cyclic permutation).
Proof For , the proof is similar to the one of the main case . For this reason, we consider further only the case where
Due to homogeneity, we may assume that . On this hypothesis,
Thus, we can write the inequality as
where and
From
it follows that g is decreasing on and increasing on , where
We have
where
We have for , where and . Since
f is convex for . Then, by PCF-Corollary, it suffices to show that for . This is true if the original inequality holds for . Thus, we need to show that
which is equivalent to the obvious inequality
□
Application 4.9 If , , , , are positive real numbers such that
then [6]
with equality for .
Proof
Write the inequality as
where and
From
it follows that g is decreasing on and increasing on , where
We have
where
We will show that for , and hence f is convex for
Indeed,
By PCF-Corollary, we only need to prove that for ; that is,
Since
it suffices to show that
This inequality is equivalent to , and the proof is completed. □
Remark 4.1
The inequality
is not true for any positive numbers , , , , , satisfying . Indeed, for , the inequality becomes
which is false for