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.
 A: 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 $(0, \pi/2)$. 
A: We may actually prove something stronger, namely
$$\forall x\in\left(0,\frac{\pi}{2}\right),\qquad \frac{\sin x}{x}+\frac{x^2}{\color{red}{6}}\geq 1.\tag{1}$$
By setting $I=\left(0,\frac{\pi}{2}\right)$, we have $0\leq \cos(t)\leq 1$ for any $t\in I$, hence by assuming $x\in I$ and integrating such inequality over the interval $(0,x)$ we get $0\leq \sin(x)\leq x$ for any $x\in I$, or $0\leq \sin(t)\leq t$ for any $t\in I$. By iterating the same argument, we get
$$ \forall x\in I,\qquad 1-\frac{x^2}{2}\leq \cos(x)\leq 1,\tag{2} $$
$$ \forall x\in I,\qquad x-\frac{x^3}{6}\leq \sin(x)\leq x\tag{3}$$
and $(1)$ is a simple consequence of $(3)$.
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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$\ds{{\sin\pars{x} \over x} + {x^{2} \over 4} > 1\,,\quad
x \in \pars{0,{\pi \over 2}}:\ ?}$

\begin{align}
\exists\ \xi\ \mid 0 < \xi < x\,,\quad 
\color{#f00}{{\sin\pars{x} \over x} + {x^{2} \over 4}} =
{4\sin\pars{x} + x^{3} \over 4x} = {4\cos\pars{\xi} + 3\xi^{2} \over 4} \color{#f00}{> 1}
\end{align}

Note that $\ds{4\cos\pars{\xi} + 3\xi^{2}}$ is a increasing function of $\ds{\xi \in \pars{0,{\pi \over 2}}}$ because

$$-4\sin\pars{\xi} + 6\xi =
6\xi\bracks{1 - {2 \over 3}\,{\sin\pars{\xi} \over \xi}} > 0\,,\quad\xi > 0
$$
$$
\mbox{such that}\quad
{4\cos\pars{\xi} + 3\xi^{2} \over 4} >
\bracks{{4\cos\pars{\xi} + 3\xi^{2} \over 4}}_{\ \xi\ \to\ 0^{+}} = 1
\qquad\xi \in \pars{0,{\pi \over 2}} 
$$
