Proving $\frac{\sin x}{x} + \frac{x^2}{4} >1$, for $x \in [0,\frac{\pi}{2}]$ 
Prove that, for every $x \in \left(0,\frac{\pi}{2}\right)$, 
  $$\frac{\sin x}{x} + \frac{x^2}{4} >1$$

I have tried using differentiation to prove that the left-hand side is strictly increasing on the interval, but no success. Please, I need a hint.

Fix $x \in (0, \pi/2)$. Using Taylor's theorem
\begin{align}
\sin x = x-\frac{1}{3!}x^3+\frac{\cos(\xi)}{5!}x^5
\end{align}
where $\xi \in (0, x)$.
Hence it follows
\begin{align}
\frac{\sin x}{x}+\frac{1}{4}x^2-1 = \left(\frac{1}{4}-\frac{1}{6}\right)x^2+\frac{\cos(\xi)}{5!}x^4>0
\end{align}
since $\cos x$ is non-negative on $(0, \pi/2)$. 
