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It's a simple problem that I propose to you this is the following :

Let $a,b,c,d,e$ be positive real numbers such that $abc=ab+bc+ca$ then we have : $$\frac{1}{da+eb}+\frac{1}{db+ec}+\frac{1}{dc+ea}\leq \frac{1}{e+d}$$

A friend tells me that there exists a very simple proof of this fact using Jensen's inequality but I don't see how...

Any hints would be appreciable.

Thanks.

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  • $\begingroup$ Assume without loss of generality that $e+d=1$ and look for convex combinations where Jensen applies. $\endgroup$ – Michal Adamaszek Nov 9 '18 at 14:00
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Using convexity of $\frac{1}{x}$ (which is Jensen in its simplest form for only two points):

  • $abc=ab+bc+ca \Leftrightarrow 1 = \sum_{cyc}\frac{1}{a}$
  • $\frac{1}{px+qy} \leq \frac{p}{x} + \frac{q}{y}$ for $x,y >0, p \in [0,1], q = 1-p$

$$\sum_{cyc}\frac{1}{da+eb} =\frac{1}{d+e}\sum_{cyc}\frac{1}{\frac{d}{d+e}a+\frac{e}{d+e}b}$$ $$\leq \frac{1}{d+e}\sum_{cyc} \left(\frac{d}{d+e}\cdot \frac{1}{a} + \frac{e}{d+e}\cdot \frac{1}{b} \right) $$$$= \frac{1}{d+e}\left(\frac{d}{d+e}\cdot 1 + \frac{e}{d+e}\cdot 1 \right) = \frac{1}{d+e} $$

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By C-S $$\sum_{cyc}\frac{1}{da+eb}\leq\frac{1}{(d+e)^2}\sum_{cyc}\left(\frac{d^2}{da}+\frac{e^2}{eb}\right)=\frac{1}{d+e}\sum_{cyc}\frac{1}{a}=\frac{1}{d+e}.$$

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