Prove $\int_x^{px}\frac{\sin^2 t}{t^2}\,\mathrm dt\leqslant\fracπ2\left(1-\frac1p\right)$ for $x\geqslant0,\ p>1$ 
Given $x\geqslant 0$ and $p>1$, define $$I (p,x) = \int_x^{px}\frac{\sin^2 t}{t^2} \,\mathrm{d}t.$$ Prove that 
  $$ I (p,x) \leqslant \frac{\pi}2\left(1-\frac1p\right).$$

So far I only figured out that $I(p,x)$ as a function of $x$, it increases in interval $[0, x_p]$, where $x_p=\dfrac{1}{p^2+p+1}$. 
But then I'm lost on how to prove $I(p,x_p) \leqslant \dfrac{\pi}2\left(1-\dfrac1p\right)$…
 A: $\def\d{\mathrm{d}}$It will be proved that$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) \tag{1}
$$
for any $0 < a < b$.
Case 1: $a \geqslant \dfrac{2}{π}$. Then$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^b \frac{1}{x^2} \,\d x = \frac{1}{a} - \frac{1}{b} = \frac{b - a}{ab}.
$$
Because$$
\frac{b - a}{ab} \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) = \frac{π}{2} \cdot \frac{b - a}{b} \Longleftrightarrow a \geqslant \frac{2}{π},
$$
then (1) holds.
Case 2: $a < \dfrac{2}{π}$ and $b \leqslant \dfrac{π}{2} \cdot \dfrac{a^2}{\sin^2 a}$.
Lemma: $\dfrac{\sin^2 x}{x^2} \leqslant \dfrac{\sin^2 a}{a^2}$ for any $x \geqslant a$.
Proof of lemma: Define $f(x) = \dfrac{\sin x}{x}$, then for any $a < x < \dfrac{π}{2}$,$$
f'(x) = \frac{1}{x^2} (x \cos x - \sin x) = \frac{\cos x}{x^2} (x - \tan x) < 0.
$$
Thus for $a < x < \dfrac{π}{2}$,$$
f(a) \geqslant f(x) > 0 \Longrightarrow \frac{\sin^2 x}{x^2} \leqslant \frac{\sin^2 a}{a^2}.
$$
For $x \geqslant \dfrac{π}{2}$, because $a < \dfrac{2}{π} < \dfrac{π}{2}$, then$$
\frac{\sin^2 a}{a^2} = (f(a))^2 \geqslant \left(f\left(\frac{π}{2}\right)\right)^2 = \frac{4}{π^2} \geqslant \frac{1}{x^2} \geqslant \frac{\sin^2 x}{x^2}.
$$
Now back to Case 2. By the lemma,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^b \frac{\sin^2 a}{a^2} \,\d x = \frac{\sin^2 a}{a^2} (b - a).
$$
Because$$
\frac{\sin^2 a}{a^2} (b - a) \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right) = \frac{π}{2} \cdot \frac{b - a}{b} \Longleftrightarrow b \leqslant \frac{π}{2} \cdot \frac{a^2}{\sin^2 a},
$$
then (1) holds.
Case 3: $a < \dfrac{2}{π}$ and $b > \dfrac{π}{2} \cdot \dfrac{a^2}{\sin^2 a}$. Now,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \int_a^{+\infty} \frac{\sin^2 x}{x^2} \,\d x = \frac{π}{2} - \int_0^a \frac{\sin^2 x}{x^2} \,\d x,
$$
and$$
\frac{π}{2} \left( 1 - \frac{a}{b} \right) \geqslant \frac{π}{2} \left( 1 - a \cdot \left(\frac{π}{2} \cdot \frac{a^2}{\sin^2 a}\right)^{-1} \right) = \frac{π}{2} - \frac{\sin^2 a}{a},
$$
thus it suffices to prove that$$
g(a) = \int_0^a \frac{\sin^2 x}{x^2} \,\d x - \frac{\sin^2 a}{a} \geqslant 0
$$
for $a < \dfrac{2}{π}$.
Because for $0 < a < \dfrac{2}{π}$,\begin{align*}
g'(a) &= \frac{\sin^2 a}{a^2} - \frac{1}{a^2} (2a \sin a \cos a - \sin^2 a)\\
&= \frac{1}{a^2} (2 \sin^2 a - 2a \sin a \cos a)\\
&= \frac{\sin 2a}{a^2} (\tan a - a) > 0,
\end{align*}
then for any $0 < a_0 < \dfrac{2}{π}$,$$
g(a_0) \geqslant \lim_{a \to 0^+} g(a) = 0 - \lim_{a \to 0^+} \frac{\sin^2 a}{a} = 0.
$$
Thus,$$
\int_a^b \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} - \int_0^a \frac{\sin^2 x}{x^2} \,\d x \leqslant \frac{π}{2} - \frac{\sin^2 a}{a} \leqslant \frac{π}{2} \left( 1 - \frac{a}{b} \right),
$$
i.e. (1) holds.
A: Actually, it is enough to remove the oscillating behaviour of the sine function through the Laplace transform. Due to a fundamental property of $\mathcal{L}$ we have
$$ \int_{a}^{b}\frac{\sin^2 x}{x^2}\,dx = \frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)+\int_{0}^{+\infty}\frac{g_b(t)-g_a(t)}{2(4+t^2)}\,dt$$
where $g_c(t)=e^{-ct}\left(t^2\cos(2c)-2t\sin(2c)\right)$. Due to the Cauchy-Schwarz inequality
$$ \int_{a}^{b}\frac{\sin^2 x}{x^2}\,dx \leq \frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)+\int_{0}^{+\infty}\frac{t(e^{-at}+e^{-bt})}{2\sqrt{4+t^2}}\,dt\leq\left(\frac{1}{a}-\frac{1}{b}\right) $$
leading to $I(p,x)\leq\color{red}{\frac{1}{x}}\left(1-\frac{1}{p}\right)$, i.e. an improvement of the given inequality for any moderately large $x$
(namely for any $x>2/\pi$).
