prove that $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$

Question

Let a,b,c,d be positive real numbers such that$\ a+b+c+d=1$, than prove that $$\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$$

This question is from the practice material of Indian Math Olympiad. I think that this will incorporate the AM-GM-HM inequality.I am not able to learn the perfect heuristic to solve this inequality.

user26486 · Accepted Answer · 2014-11-08T17:11:25.207

6

By AM-GM:

$ab\le \left(\frac{a+b}{2}\right)^2=\frac{(a+b)^2}{4}$

Thus

$$\sum_{\text{sym}}\frac{ab}{a+b}\le \sum_{\text{sym}}\frac{a+b}{4}=\frac{3}{4}$$

Equality is achieved iff $a=b=c=d$.

edited Nov 08 '14 at 17:11

answered Nov 08 '14 at 17:05

user26486

11,331

@RishabhGoswami $\sum_{\text{sym}}(a+b)=3(a+b+c+d)=3$. – user26486 Nov 08 '14 at 17:14
I.e. iff $a=b=c=d=\frac{1}{4}$. – user26486 Oct 15 '17 at 15:04

prove that $\frac{ab}{a+b}+\frac{ac}{a+c}+\frac{ad}{a+d}+\frac{bc}{b+c}+\frac{bd}{b+d}+\frac{cd}{c+d}\le \frac{3}{4}$

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