The following is part of exercise 6.26.21 from Tom Apostol's Calculus Volume 1. I wonder if my proof is correct and if there is a simpler alternative proof.
Prove the following by examining the sign of the derivative of an appropriate function: $$ \frac{2}{\pi}x \lt \sin{x} \qquad \text{if} \qquad 0 \lt x \lt \frac{\pi}{2} \tag{1}\label{1} $$
Let $ f(x)=\sin{x}-\frac{2}{\pi}x $, $ 0 \le x \le \frac{\pi}{2} $ then
$$ f(0)=f\left(\frac{\pi}{2}\right)=0 \tag{2}\label{2} $$
and
$$ f''(x)=-\sin{x} \lt 0 \tag{3}\label{3} $$
From $ \eqref{2} $ and $ \eqref{3} $ we know that $ f $ has a maximum at exactly one point, this together with $ \eqref{2} $ proves $ \eqref{1} $.
Your proof is fine: the concavity of the sine function over $\left(0,\frac{\pi}{2}\right)$ gives the wanted inequality in a straightforward way. Anyway, if you like to kill flies with hydrogen bombs, you may consider that: $$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2 \pi^2}\right) $$ hence if $x\in\left(0,\frac{\pi}{2}\right)$ we have: $$ \frac{\sin x}{x}> \prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right) = \prod_{n\geq 1}\frac{2n-1}{2n}\cdot\frac{2n+1}{2n} $$ where the RHS is the reciprocal of the Wallis product, i.e. $\frac{2}{\pi}$ as wanted.