Hello I want to prove that $$4\sum_{cyc} \frac{a}{(a+b)^2}\le\frac1a+\frac1b+\frac1c$$ for all $a,b,c>0$. I try multiplying by all the denominators and expanding but I get over 30 terms. Is there a nice proof?


1 Answer 1


Hint: $(a+b)^2 \ge 4ab$, so $$\frac{a}{(a+b)^2}\le\cdots$$

  • $\begingroup$ What a easy proof $\endgroup$
    – MafPrivate
    Commented Nov 26, 2019 at 14:43

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