$x,y,z$ are positive real numbers such that $$xyz=1$$ Prove that $\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$. I have no idea how to solve this problem. I've tried Engel form of Cauchy inequality, but I get $x^2+y^2+z^2$ in denominator, and from condition $xyz=1$ I only find $x^2+y^2+z^2\geq3..$


$$\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$$

Since $xyz=1$ we substitute, $x=\frac{a}{b} , y= \frac{b}{c} , z= \frac{c}{a} $

Now, by Engel form(also called Titu lemma)

$$ \sum_{cyc}\frac {a^2}{c^2 + ab }\ge\frac {(a^2 + b^2 +c^2)^2}{a^2b^2 + b^2c^2 + c^2a^2+a^3b + b^3c + c^3a} $$

So it remains to prove

$$ 2(a^2 +b^2+c^2)^2\ge 3\sum_{cyc}a^2b^2 + 3\sum_{cyc}a^3b $$ $$ 2a^4+2b^4+2c^4 +a^2b^2+b^2c^2+a^2c^2 \ge 3(a^3b+b^3c+c^3a) $$ Which is obvious from,

$$ a^4 + b^4 +c^4 \ge a^3b + b^3c + c^3a $$

$$ \sum_{cyc}(a^4 + a^2b^2)\ge \sum_{cyc}2a^3b $$

  • $\begingroup$ do you have an easier way? $\endgroup$ – RE60K Sep 20 '14 at 15:16
  • $\begingroup$ By rearrangement inequality :) $\endgroup$ – Shivang jindal Sep 20 '14 at 18:58

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