$f(x)=x^2-x\sin x, x\in [0,\pi/2]$
I reduced it to finding that $f'(x)=2x-\sin x- x\cos x>0 \space\forall x \in [0,\pi/2]$
I then simplified $f'(x)=2x-\sqrt{1+x^2}\sin({x+\arctan x})$
Now $f'(x)>2x-\sqrt{1+x^2}$ which is $>0 \forall x\in[1/\sqrt3,\pi/2]$
How to prove $f'(x)>0$for the intervel $[0,1/\sqrt3]$?
Since $f'(x)=x-\sin(x)+x[1-\cos(x)]>0$ holds for all $x\in(0,\pi/2)$, the function $f$ is strictly increasing on $[0,\pi/2]$. This implies that it attains its minimum at $x=0$ and its maximum at $x=\pi/2$.