Proving an inequality of trigonometric functions I am trying to prove that $$(\cos x)^{\cos x} > (\sin x)^{\sin x}$$ for $x \in \left(0, \frac{\pi}{4}\right)$.
I have plotted the graph of $y=(\cos x)^{\cos x} - (\sin x)^{\sin x}$ and see that my conjecture is supported.
Therefore, I tried to prove an equivalent statement $$ \cos x  \ln  \cos x> \sin x \ln \sin x$$.
I tried to use Calculus to prove the statement.
Let $$f(x)= \cos x  \ln  \cos x-\sin x \ln \sin x$$
then $$f'(x)=-\sin x\ln \cos x -\cos x \ln \sin x -\sin x - \cos x.$$
From here I don't know how to proceed.
Any help to prove the inequality will be much appreciated!

We need to prove that $f(x)>0$, where
$$f(x)=x\ln{x}-\frac{1}{2}\sqrt{1-x^2}\ln(1-x^2)$$
for $\frac{1}{\sqrt2}<x<1$.
Indeed, $$f'(x)=\ln{x}+1+\frac{x}{2\sqrt{1-x^2}}\left(2+\ln(1-x^2)\right);$$
$$f''(x)=\frac{x\ln(1-x^2)+2(1-x^2)(x+\sqrt{1-x^2})}{2x\sqrt{(1-x^2)^3}}.$$
Let $$g(x)=\ln(1-x^2)+\frac{2(1-x^2)\left(x+\sqrt{1-x^2}\right)}{x}.$$
Thus, $$g'(x)=-\frac{2x}{1-x^2}-2\sqrt{1-x^2}\left(2+\frac{1}{x^2}\right)-4x<0.$$
$\lim\limits_{x\rightarrow1^-}f''(x)=-\infty$ and $f''\left(\frac{1}{\sqrt2}\right)>0$, 
which says that there is unique $x_1\in\left(\frac{1}{\sqrt2},1\right)$, for which $f''=0$.
Indeed, $x_1=0.8197...$ and in this point we have a maximum of $f'$.
We have $f'\left(\frac{1}{\sqrt2}\right)>0$, which says that on $\left[\frac{1}{\sqrt2},x_1\right]$ $f'$ increasing. Also $f\left(\frac{1}{\sqrt2}\right)=0$.
Thus, our inequality is proven for $\frac{1}{\sqrt2}<x\leq x_1$.
But on $\left[x_1,1\right)$ $f$ is a concave function, $f(x_1)=0.156...>0$ and $\lim\limits_{x\rightarrow1^-}f(x)=0$,
which ends the proof.
