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

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

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

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 – Sualeh Asif · 10 months, 3 weeks ago

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– Ameya Daigavane · 10 months, 3 weeks ago

Well, proving the Vasc Inequality might be slightly more difficult.Log in to reply

It does not yeild to most classical inequalities! – Sualeh Asif · 10 months, 3 weeks ago

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– Ameya Daigavane · 10 months, 3 weeks ago

Yes, I know. I doubt it can be quoted in contests either.Log in to reply

AoPS:

I found this amazingly elegant proof on\( \left(a^2 + b^2 + c^2\right)^2 - 3\left(a^3b + b^3c + c^3a\right) \\ = \frac {1}{2}\left( \left( a^{2} - 2ab + bc - c^{2} + ca\right) ^2 + \left(b^{2} - 2bc + ca - a^{2} + ab\right)^2 + \left( c^{2} - 2ca + ab - b^{2} + bc\right)^2 \right) \geq 0 \) – Ameya Daigavane · 10 months, 3 weeks ago

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– Sualeh Asif · 10 months, 3 weeks ago

This is basically the only relatively clean proof!Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Awesome!!! (speechless)Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Yep! Titu and Vasc killed this... =DLog in to reply

@Gurīdo Cuong – Ms Ht · 10 months, 3 weeks ago

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@MS HT @Sharky Kesa @ZK LIn – Abdur Rehman Zahid · 10 months, 3 weeks ago

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– Sharky Kesa · 10 months, 3 weeks ago

You really want my long, expanded solution?Log in to reply

– Abdur Rehman Zahid · 8 months, 4 weeks ago

I still haven't seen your long and expanded solution Sharky :PLog in to reply

– Sualeh Asif · 8 months, 4 weeks ago

It isnt very difficult that way. Try yourself expanding the whole inequality,without any denominators. Then work with AM-GM and Shur. (And post the expanded stuff)Log in to reply

– Abdur Rehman Zahid · 10 months, 3 weeks ago

Yes please.Log in to reply

– Nihar Mahajan · 10 months, 3 weeks ago

Yes :PLog in to reply