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If $x,y,z$ are positive real numbers such that $xyz \geqslant 1$ prove: $$\dfrac{x^3+y^3}{x^2+xy+y^2}+\dfrac{y^3+z^3}{y^2+yz+z^2}+\dfrac{x^3+z^3}{x^2+xz+z^2} \geqslant 2$$

I have tried with Hölder's inequality, but it is not working. Can you help, please?

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Since: $$\frac23x^3+\frac13y^3\ge x^y\text{ or }\frac23y^3+\frac13x^3\ge y^2x$$ So: $$\large x^3+y^3\ge\frac13x^3+\frac23x^2y+\frac23xy^2+\frac13y^3=\frac13(x+y)(x^2+xy+y^2)$$ Hence: $$\large \sum_{cyc}\frac{x^3+y^3}{x^2+xy+y^2}\ge\sum_{cyc}\frac{x+y}{3}=\frac23(x+y+z)\ge2\sqrt[3]{xyz}=2$$

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  • $\begingroup$ How do you prove $$x^3+y^3 \geqslant x^2y+xy^2$$? $\endgroup$ – chaos Sep 15 '14 at 10:33
  • $\begingroup$ That's a very basic inequality. If you want to be able to deal with such problems you'll have to learn the basics of inequalities first. $\endgroup$ – Matheo Sep 15 '14 at 11:24
  • $\begingroup$ @chaos see the edit. $\endgroup$ – RE60K Sep 15 '14 at 15:51

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