# Inequality using Jensen

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.

• Assume without loss of generality that $e+d=1$ and look for convex combinations where Jensen applies. – Michal Adamaszek Nov 9 '18 at 14:00

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 $$x,y >0, p \in [0,1], q = 1-p$$
$$\sum_{cyc}\frac{1}{da+eb} =\frac{1}{d+e}\sum_{cyc}\frac{1}{\frac{d}{d+e}a+\frac{e}{d+e}b}$$ $$\leq \frac{1}{d+e}\sum_{cyc} \left(\frac{d}{d+e}\cdot \frac{1}{a} + \frac{e}{d+e}\cdot \frac{1}{b} \right)$$$$= \frac{1}{d+e}\left(\frac{d}{d+e}\cdot 1 + \frac{e}{d+e}\cdot 1 \right) = \frac{1}{d+e}$$
By C-S $$\sum_{cyc}\frac{1}{da+eb}\leq\frac{1}{(d+e)^2}\sum_{cyc}\left(\frac{d^2}{da}+\frac{e^2}{eb}\right)=\frac{1}{d+e}\sum_{cyc}\frac{1}{a}=\frac{1}{d+e}.$$