How to prove: $\sin^2 x\ge \ln(1+x^2),~~x\in[0,1]$
My attempt:
I try to use series to prove it.
$$\begin{align}\sin^2x=\frac{1-\cos(2x)}2&=\frac{1}2\left(1-\sum_{n=0}^\infty\frac{(-1)^n\cdot4^n}{(2n)!}x^{2n}\right)\\ \\ \sin^2x&=\sum_{n=1}^\infty\frac{(-1)^{n+1}\cdot2^{2n-1}}{(2n)!}x^{2n}\\ \\ \ln(1+x^2)&=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}x^{2n}\\ \\ \sin^2x-\ln(1+x^2)&=\sum_{n=1}^\infty(-1)^{n+1}\left(\frac{2^{2n-1}}{(2n)!}-\frac{1}{n}\right)x^{2n}\end{align}$$
I know when $n$ is large, then $\frac{2^{2n-1}}{(2n)!}<\frac{1}{n}$, but there is an alternating term $(-1)^{n+1}$. Also the behavior for $x\to0$ and $x\to 1$ is different when compare the difference between this two series. So I stuck here. How can I proceed on this series method? Or are there other ways to prove it? Thank you!
Here is the plot for $y=\sin^2x-\ln(1+x^2)$, $y_1=y',~y_2=y''$
