Find $\sin^5 x + \cos^5 x$ 
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$

 My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x \cos^3x+\sin^4x)\\
&=(\sin x+\cos x)[(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x \cos x-\sin^2x\cos^2x]\\
&=(\sin x+\cos x)[1-\sin x\cos x-\sin^2x\cos^2x]\\
&=\frac{1}{4}(\sin x+\cos x)(4-2\sin 2x- \sin^2 2x)\\
\end{align*} I don't know what next?

I wanted to develop the idea expressed in comments that LHS is bounded below by $\cos(1)$ while RHS is bounded above by $\cos(1)$ and that we get equality for $x=0\text{ or }\frac{\pi}2$ only.

*

*The LHS part and equality case is not too complicated to prove:

$\require{cancel}\begin{align}f(x)
&=\Big(\cos(x)+\sin(x)\Big)\Big(\overbrace{1}^{\cos(x)^2+\sin(x)^2}-\sin(x)\cos(x)\Big)\\
&=\cos(x)^3+\cancel{\cos(x)\sin(x)^2}-\cancel{\sin(x)\cos(x)^2}+\cancel{\sin(x)\cos(x)^2}+\sin(x)^3-\cancel{\sin(x)^2\cos(x)}\\
&=\cos(x)^3+\sin(x)^3\end{align}$
But notice that $\forall n\in\mathbb N$ since $|\sin(x)|\le 1\implies |\sin(x)|^n\le |\sin(x)|^2$ and same for cosine we get,
$\Big|f(x)\Big|=\Big|\sin(x)^3+\cos(x)^3\Big|\le \Big|\sin(x)\Big|^3+\Big|\cos(x)\Big|^3\le \sin(x)^2+\cos(x)^2=1$
Notice also that $|f(x)|$ has period $\pi$ and that $\cos\searrow$ on $[0,\pi]$
Therefore (and since RHS undefined where $f(x)$ is negative):
$$\cos\left(\sqrt{|f(x)|}\right)\ge \cos(1)\quad\text{and}\quad LHS\ge \cos(1)$$
The equality case is for extremal points.
So let search for critical points $f'(x)=3\sin(x)\cos(x)(\sin(x)-\cos(x))=0\iff x=0,\frac{\pi}2,\frac{\pi}4$ over the $[0,\pi]$ interval, among which only $0,\frac{\pi}2$ lead to $f(x)=1$.


*

*However, I faced a wall when I tried to prove the RHS part, so it seems basilik's approach despite looking more complicated at first sight, is eventually more effective.

Does anyone attempted this approach too ?
