# Inequality $a^{2b}+b^{2a}\leq \cos(ab)^{(a-b)^2}$

It's well know that we have (inequality of Vasile Cirtoaje) :

Let $$a,b>0$$ such that $$a+b=1$$ then : $$a^{2b}+b^{2a}\leq 1$$

The following is a add of mine :

Let $$a,b>0$$ such that $$a+b=1$$ then : $$a^{2b}+b^{2a}\leq \Big(\cos(ab)\Big)^{(a-b)^2}$$

I try the following inequalities :

$$a^{2b}\leq \beta \Big(\cos(ab)\Big)^{(a-b)^2}$$ $$b^{2a}\leq (1-\beta) \Big(\cos(ab)\Big)^{(a-b)^2}$$

And take log on both side but I'm stuck on this way.

I try furthermore to use convexity but I gave up quickly.

Finally I try Bernoulli's inequality but it's too weak .

Any helps are very appreciated .

Thanks a lot for your time and patience .

• Just to clarify, do you mean $\cos (ab)^{(a-b)^2}=(\cos (ab))^{(a-b)^2}$? Otherwise clearly it is not true, right? Commented Feb 24, 2020 at 20:55
• @Rob Yes I mean this .Let me edit .Thanks for the remark . Commented Feb 25, 2020 at 11:54

Due to symmetry, assume that $$a \ge b$$.
By using $$a = 1- b$$ and $$b\in (0, \frac{1}{2}]$$, the desired inequality is written as $$(1-b)^{2b} + b^{2 - 2b} \le (\cos (b - b^2))^{(1-2b)^2}.$$

Let us first prove the case when $$b\in (0, \frac{1}{3}]$$.

We give the following auxiliary results (Facts 1 through 3). Their proof is not difficult and hence omitted.

Fact 1: $$(1-x)^r \le 1 - rx + \frac{1}{2}(r^2-r)x^2$$ for $$x\in (0, \frac{1}{2}]$$ and $$0 < r \le 1$$.

Fact 2: $$\cos y \ge 1 - \frac{1}{2}y^2$$ for $$y \in (0, +\infty)$$.

Fact 3: From Bernoulli inequality, $$(1-x)^r \ge 1 - rx$$ for $$r \ge 1$$ and $$0 < x < 1$$.

From Facts 2 and 3, we have for $$b \in (0, \frac{1}{2}]$$, \begin{align} &(\cos (b - b^2))^{(1-2b)^2} \\ \ge\ & [1 - \tfrac{1}{2}(b - b^2)^2]^{(1-2b)^2}\\ = \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1}[1 - \tfrac{1}{2}(b - b^2)^2]^{1+(1-2b)^2}\\ \ge \ & [1 - \tfrac{1}{2}(b - b^2)^2]^{-1} \Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big). \end{align} From Fact 1, by letting $$x = b, r = 2b$$, we have for $$b \in (0, \frac{1}{2}]$$, $$(1-b)^{2b} \le (1-b)(1+b - b^2 - 2b^3).$$

Thus, it suffices to prove that for $$b\in (0, \frac{1}{3}]$$, $$(1-b)(1+b - b^2 - 2b^3) + b^{2-2b} \le [1 - \tfrac{1}{2}(b - b^2)^2]^{-1} \Big(1 - \tfrac{1}{2}(b - b^2)^2[1 + (1-2b)^2] \Big)$$ or $$\frac{b^2(2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3)}{-b^4+2 b^3-b^2+2}-b^{2-2b}\ge 0$$ or $$\frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2}-b^{-2b}\ge 0$$ or $$\ln \frac{2 b^6-5 b^5-2 b^4+15 b^3-19 b^2+8 b+3}{-b^4+2 b^3-b^2+2} + 2b \ln b \ge 0.$$ It is not difficult with computer. Indeed, denote LHS by $$f(b)$$. One can prove that $$f''(b) > 0$$ for $$b\in (0, \frac{1}{3}]$$. Then one can prove that $$f'(b) < 0$$ for $$b\in (0, \frac{1}{3}]$$. Omitted.