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Let $ f$ be a strictly increasing, continuous function mapping $ I=[0,1]$ onto itself. Prove that the following inequality holds for all pairs $ x,y \in I$: $$ 1-\cos (xy) \le\int_0^xf(t) \sin {(tf(t))}dt + \int_0^y f^{-1}(t) \sin{(tf^{-1}(t))} dt .$$

For this problem I can want use Young inequality

so let $$g(t)=f(t)\sin{(tf(t))},g(0)=0$$

But $g(t)$ is strictly increasing function ? and $g^{-1}(t)=?$

so I can't use Young inequality to slove this problem.Maybe can use other methods? Thank you

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Denote \begin{align} T_1 &:=\{(u,v)\,|\,0\leq u\leq x,\,0\leq v\leq y\},\\ T_2 &:=\{(u,v)\,|\,0\leq u\leq x,\,0\leq v\leq f(u)\},\\ T_3 &:=\{(u,v)\,|\,0\leq v\leq y,\,0\leq u < f^{-1}(v)\}. \end{align} Then $T_1\subset T_2\cup T_3$, $T_2\cap T_3=\emptyset$ and all the three sets are Borel sets. So for every $g:[0,1]^2\to\mathbf{R}$, $g\geq 0$, Lebesgue-integrable function we have $$ \int_0^y\int_0^x g(u,v)\,du\,dv \leq \int_0^x\int_0^{f(u)} g(u,v)\,dv,du+\int_0^y\int_0^{f^{-1}(v)} g(u,v)\,du\,dv. $$ Taking $$ g(x,y):=\frac{\partial^2(1-\cos(xy))}{\partial y\partial x}=\sin(xy)+xy\cos(xy)\geq 0 $$ we obtain $$ 1-\cos (xy) \le\int_0^xf(u) \sin {(u\cdot f(u))}du + \int_0^y f^{-1}(v) \sin{(v\cdot f^{-1}(v))} dv. $$

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From a geometric perspective, we need to prove that the right hand side is larger than than the area of the rectangle $xy$, and hench larger than $1-\cos(xy)$.If the integrands were simply $f(t)$ and $f'(t)$, the diagram below would suffice. Since $\sin$ is monotonous as well as $t f(t)$, you should be able to formulate a similar argument even though the integrands are no longer true inverses of each other. $$$$ $\quad\quad\quad\quad\quad\quad\quad\quad$enter image description here

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