Let us consider the function $$I(p):= \frac {\Gamma(2-p)\Gamma(3p)}{(p\Gamma(p))^2} $$ on the interval $(0,1),$ where $\Gamma(x)$ denotes the gamma function. How to prove its convexity there?
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3By the way, are asking for help or you want to tell us what to do? – Mhenni Benghorbal Jun 26 '13 at 05:03
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By the way, where did this question come from? – Mhenni Benghorbal Jun 26 '13 at 05:34
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Using the identity $p\Gamma(p)=\Gamma(p+1)$ and the expression of $\Gamma$ as a Weierstrass product, one gets $$ I(p)=\prod_{n=0}^{+\infty}i_n(p),\qquad i_n(x)=\frac{(x+1+n)^2}{(2-x+n)(3x+n)}. $$ Using logarithmic derivatives, one sees that each function $i_n$ is decreasing on $(0,\frac12)$ and increasing on $(\frac12,1)$. Its (usual) second derivative is $$ i''_n(x)=(2n+3)\left(\frac3{(3x+n)^3}+\frac1{(2-x+n)^3}\right), $$ hence each $i_n$ is positive and convex on $(0,1)$. Thus, their product $I$ is convex and decreasing on $(0,\frac12)$ and is convex and increasing on $(\frac12,1)$.
Every function continuous on $(0,1)$, convex and decreasing on $(0,\frac12)$, and convex and increasing on $(\frac12,1)$, is convex on $(0,1)$ because every chord from $x$ in $(0,\frac12)$ to $x'$ in $(\frac12,1)$ lies above the chords from $x$ to $\frac12$ and from $\frac12$ to $x'$. This proves that $I$ is convex on $(0,1)$.
