Prove $\sin^2(x)<\sin(x^2)$ for $0<x<\sqrt{\frac{\pi}{2}}$

I'm trying to prove $\sin^2(x)<\sin(x^2)$ for $0<x<\sqrt{\frac{\pi}{2}}$.

Attempt: This is equivalent to showing $f(x)=\sin(x^2)-\sin^2(x) = \sin(x^2)-\left(\frac{1-\cos(2x)}{2}\right)>0$. Since $f(0)=0$, if we show $f$ is increasing we are done. $f'(x) = 2x\cos(x^2)-\sin(2x)$ but I can't see why it's greater than zero.

Solution 1:

Finishing ellya's proof: For $x>1$ we have $$\sin^2(x)< \sin(x) < \sin(x^2)$$ Since $\sin(x)$ is increasing in our domain. For $x\le1$: $$\cos(x^2)\ge\cos(x), x>\sin(x)\implies 2x\cos(x^2)>2\sin(x)\cos(x)=\sin(2x)$$

Solution 2:

My thoughts at the moment are to take a step back from what you have put, and see that on $0\lt x\lt\sqrt{\frac{\pi}{2}}$ , $\sin(x)\lt1$ so here $\sin^2(x)\lt\sin(x)$, now it is a question of showing that $\sin(x)\lt \sin(x^2)$