<s>[INST] Getting strict inequality in $|f(x)|\le \frac 1x$, $x\gt 0$, where $f(x)=\int_x^{x+1} \sin t^2 \, dt$ Given that $f(x)=\int_x^{x+1} \sin t^2 \, dt$
Substituting $u=t^2$ gives:
$f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du$
It follows that $|f(x)|\le\frac 1{2x}+\frac 1{2(x+1)}+\frac 1{2x}-\frac 1{2(x+1)}=\frac 1x$
But how to refute the possibility that $|f(x)|=\frac 1x$ for some $x\gt 0$?
I tried to prove the strict inequality using contradiction as follows:
Suppose on the contrary that there exists a $y\gt 0$ such that $|f(y)|=\frac 1y$, then $|f(y)|=\frac 1y=|\int_y^{y+1}\sin t^2\,dt|\le1\implies y\ge 1$
$y\ne 1$ as $y=1\implies |f(1)|=1=\int_1^2\sin t^2\, dt$, which is not true as RHS is approximately $0.50$. So we must have $y\gt 1$.
Let $g(x)=\frac 1x-f(x)$ and then I try to see if $g$ is monotonic on $(1,\infty)$ or not  but $g'(x)=-\frac 1{x^2}+\sin x^2-\sin(x+1)^2\ge-1+\sin x^2-\sin (x+1)^2$.
I am stuck here. Please help. Thanks.
 [/INST] You correctly derived that
$$
f(x)=\int_{x^2}^{(x+1)^2}\frac{\sin u}{2\sqrt u}\,du=\frac{\cos x^2}{2x}-\frac{\cos (x+1)^2}{2(x+1)}-\int_{x^2}^{(x+1)^2}\frac{\cos u}{4u^{3/2}}\, du \, .
$$
It follows that
$$
|f(x)| \le\frac{|\cos x^2|}{2x}+\frac{|\cos (x+1)^2|}{2(x+1)}+\int_{x^2}^{(x+1)^2}\frac{|\cos u|}{4u^{3/2}}\, du  \\
\overset{(*)}{<} \frac{1}{2x}+\frac{1}{2(x+1)}+\int_{x^2}^{(x+1)^2}\frac{1}{4u^{3/2}}\, du = \frac 1x \, .
$$
We have a strict inequality at $(*)$ because $|\cos u| = 1$ does not hold for all $u \in [x, x+1]$.
</s>