# Prove that $a^{\cos(b)^2}-b^{\cos(a)^2}\leq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$ and the reverse inequality .

It'a problem of my own :

Let $$a\geq b>0$$ such that $$a+b=\frac{\pi}{2}$$ then we have : $$a^{\cos(b)^2}-b^{\cos(a)^2}\leq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$$ Let $$b\geq a>0$$ such that $$a+b=\frac{\pi}{2}$$ then we have : $$a^{\cos(b)^2}-b^{\cos(a)^2}\geq \frac{4}{\pi}(1-\frac{\pi}{2})b+\frac{\pi}{2}-1$$

For the first the equality case is when $$b=0$$ or $$b=\frac{\pi}{4}$$ for the second when $$b=\frac{\pi}{4}$$ or $$b=\frac{\pi}{2}$$

The main remark is : the line $$f(x)=\frac{4}{\pi}(1-\frac{\pi}{2})x+\frac{\pi}{2}-1$$ is a chord of the curve defines by the function $$g(x)=\Big(\frac{\pi}{2}-x\Big)^{\cos^2(x)}-\Big(x\Big)^{\sin^2(x)}$$

So my idea was to derivate the function $$g(x)$$ we obtain :

$$g'(x)= (\frac{\pi}{2} - x)^{\cos^2(x)} \Big(-\frac{\cos^2(x)}{(\frac{\pi}{2} - x)} - 2 \log\Big(\frac{\pi}{2} - x\Big) \sin(x) \cos(x)\Big) - x^{\sin^2(x)} \Big(\frac{\sin^2(x)}{x} + 2 \log(x) \sin(x) \cos(x)\Big)$$

See that $$g'(x)<0$$ on the interval $$[0,\frac{\pi}{2}]$$ and apply the mean value theorem but I'm stuck after this.

If you have nice idea it would be cool.

Ps: Is there a symmetry with respect to the line ($$f(x)$$)?If yes we can cut in two the problem.

Thanks for sharing your time and knowledge .

put $$x=\dfrac{\pi}{4}+a$$
# $$h(a)=(\frac{\pi}{4}-a)^{\frac{1}{4}(\cos{a}-\sin{a})}+(\frac{\pi}{4}+a)^{\frac{1}{4}(\sin{a}+\cos{a})}$$
$$=i(a)+i(-a)$$
$$i''(x)$$ is increasing in the area $$(-\frac{\pi}{4},0)$$ and $$i'(0)\geq-1$$, h(x) is the shape like which add two parabola.