Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{a^3c}{(a+c)(b+c)}+\frac{b^3a}{(b+a)(c+a)}+\frac{c^3b}{(c+b)(a+b)}\geq\frac{3}{4}$$ I tried C-S and BW. It does not help.
If $abc=1$ so $\sum\limits_{cyc}\frac{a^3c}{(a+c)(b+c)}\geq\frac{3}{4}$
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inequality
contest-math
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0Ok, that does not work, sorry. Since the LHS is similar to a Lagrange interpolating polynomial, have you tried partial fraction decomposition? – 2017-02-01
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0setting $$a=x/y,b=y/z,c=z/x$$ we get an inequality in $x,y,z$ after this BW works – 2017-04-08
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0Sonnhard I tried BW. – 2017-04-08