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Does anyone know a simple proof of the following Prékopa-Leindler style inequality:

If we have $f_1,f_2,g_1,g_2$ strictly positive functions on $\mathbb{R}$ such that, for any $x_1,x_2 \in \mathbb{R}$, one has $$f_1(x_1)^2 f_2(x_2)^3 \leq g_1(x_1)^2g_2(x_2)^3 $$ then $$ \left(\int_\mathbb{R} f_1\right)^2\left(\int_\mathbb{R}f_2\right)^3 \leq \left(\int_\mathbb{R}g_1\right)^2\left(\int_\mathbb{R} g_2\right)^3.$$

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1 Answer 1

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Is not it obvious (unlike Prékopa-Leindler)?

We are given that for all $x_1,x_2$ we have $(f_1/g_1)^2 (x_1)\leqslant (g_2/f_2)^3(x_2)$, thus there exists $c>0$ such that $(f_1/g_1)^2 (x_1)\leqslant c^6\leqslant (g_2/f_2)^3(x_2)$, i.e. $f_1(x)\leqslant c^3 g_1(x)$, $c^2 f_2(x)\leqslant g_2(x)$ for all $x$, we integrate these pointwise inequalities, take appropriate powers of integrals and multiply.

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  • $\begingroup$ Hi ! Thx for your answer ! Then if you mix the right-hand side of the pointwise inequality as $g_1(x_1 + x_2)^2 g_2(x_1 - x_2)^3$, i guess it is not trivial anymore.. $\endgroup$
    – Anthony
    Commented Apr 18 at 8:25
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    $\begingroup$ At least, not so trivial. But if you want to ask this, please make a new question with all details. $\endgroup$ Commented Apr 18 at 8:42

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