\[\dfrac{x^3}{z^3+x^2y}+\dfrac{y^3}{x^3+y^2z}+\dfrac{z^3}{y^3+z^2x}\geq \dfrac{3}{2}\]

Given that \(x,y\) and \(z\) are positive reals, prove the inequality above.

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## Comments

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TopNewestHere is a short solution I found:

\[\sum_{cyc} \dfrac{x^4}{xz^3+x^3z} \geq \dfrac{(x^2+y^2+z^2)^2}{2(xz^3+x^3y+y^3z)}\geq \dfrac{3}{2}\]

Hence it suffices to prove that

\[(x^2+y^2+z^2)^2\geq 3(xz^3+x^3y+y^3z)\]

Which is the famous Vasc's inequality

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Well, proving the Vasc Inequality might be slightly more difficult.

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To be exact Tremendously difficult

It does not yeild to most classical inequalities!

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AoPS:

I found this amazingly elegant proof on\( \left(a^2 + b^2 + c^2\right)^2 - 3\left(a^3b + b^3c + c^3a\right) \\ = \frac {1}{2}\left( \left( a^{2} - 2ab + bc - c^{2} + ca\right) ^2 + \left(b^{2} - 2bc + ca - a^{2} + ab\right)^2 + \left( c^{2} - 2ca + ab - b^{2} + bc\right)^2 \right) \geq 0 \)

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Yep! Titu and Vasc killed this... =D

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@Gurīdo Cuong

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@MS HT @Sharky Kesa @ZK LIn

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You really want my long, expanded solution?

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I still haven't seen your long and expanded solution Sharky :P

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Yes please.

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Yes :P

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