TITLE: How to prove that $\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$ for $p >1, x\ge0$
QUESTION [11 upvotes]: Show that for $p>1$ and $x \ge 0$:
$$\dfrac{2}{\pi}\int_{x}^{px}\left(\dfrac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$$

My idea is to use $$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\cdots $$

REPLY [10 votes]: This is quite a difficult problem, and I found it very enjoyable.  Here is the solution I found:  
First, we give some simple bounds when $x$ is large, or $px$ is small.  If $x\geq\frac{2}{\pi},$ then by using the bound $|\sin(t)|\leq1$,
we have that
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\int_{x}^{px}\frac{1}{t^{2}}dt=\frac{2}{\pi x}\left(1-\frac{1}{p}\right)\leq1-\frac{1}{p}.
$$
Similarly, if $px\leq\frac{\pi}{2}$, then since $\frac{\text{sin}(t)}{t}\leq1$,
it follows that 
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\left(px-x\right)=\frac{2xp}{\pi}\left(1-\frac{1}{p}\right)\leq\left(1-\frac{1}{p}\right).
$$
Now, assume that $0\leq x\leq\frac{2}{\pi}$, and that $px\geq\frac{\pi}{2}$.
Then notice that 
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt=1-\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt-\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt
$$
since $\int_{0}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt=\frac{\pi}{2}$.
We will now find a bound on the other two terms. Working over an interval
of length $\pi$, by pulling out a lower bound for $\frac{1}{t^{2}}$,
we have that for any $y$ 
$$
\int_{y}^{y+\pi}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{\left(y+\pi\right)^{2}}\int_{0}^{\pi}\sin^{2}(t)dt\geq\frac{\pi}{2}\int_{y+\pi}^{y+2\pi}\frac{1}{t^{2}}dt,
$$
and so 
$$
\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\int_{px+\pi}^{\infty}\frac{1}{t^{2}}dt=\frac{1}{px+\pi}.
$$
Since the function $\frac{\sin(t)}{t}$ is monotonically decreasing
on the interval $\left[0,\frac{2}{\pi}\right],$ it follows that for
$x\leq\frac{2}{\pi}$ we have 
$$\frac{1}{x}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\int_{0}^{\frac{2}{\pi}}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\cdot\frac{5}{3\pi},$$
and hence 
$$
\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{5x}{3\pi}.
$$ 
Now, notice that since $px\geq\frac{\pi}{2},$ and $p>1$, by plugging them in directly, we have that 
$$
\frac{5\left(xp\right)^{2}}{3\pi}+\frac{2}{3}px+p-\pi>\frac{5\pi}{12}+\frac{\pi}{3}+1-\pi=1-\frac{\pi}{4}>0.
$$ 
Rearranging the above by dividing through by both $(px+\pi)$  and $p$, we obtain the inequality
$$\frac{5}{3\pi}x+\frac{1}{px+\pi}>\frac{1}{p},$$ 
for $px\geq\frac{\pi}{2}$, and $p>1$. It then follows that
$$
\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt+\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{p},
$$
 for $x\leq\frac{2}{\pi},$ and $px\geq\frac{\pi}{2}$, and hence
we have shown that for all $x\geq0$, and all $p>1$, 
$$
\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq1-\frac{1}{p},
$$
as desired.