NMTC Problem

x3z3+x2y+y3x3+y2z+z3y3+z2x32\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,yx,y and zz are positive reals, prove the inequality above.

Note by Abdur Rehman Zahid
3 years, 8 months ago

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Here is a short solution I found:

cycx4xz3+x3z(x2+y2+z2)22(xz3+x3y+y3z)32\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

(x2+y2+z2)23(xz3+x3y+y3z)(x^2+y^2+z^2)^2\geq 3(xz^3+x^3y+y^3z)

Which is the famous Vasc's inequality

Sualeh Asif - 3 years, 8 months ago

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

Ameya Daigavane - 3 years, 8 months ago

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To be exact Tremendously difficult

It does not yeild to most classical inequalities!

Sualeh Asif - 3 years, 8 months ago

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@Sualeh Asif Yes, I know. I doubt it can be quoted in contests either.

Ameya Daigavane - 3 years, 8 months ago

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@Sualeh Asif I found this amazingly elegant proof on AoPS:
(a2+b2+c2)23(a3b+b3c+c3a)=12((a22ab+bcc2+ca)2+(b22bc+caa2+ab)2+(c22ca+abb2+bc)2)0 \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 - 3 years, 8 months ago

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@Ameya Daigavane Awesome!!! (speechless)

Nihar Mahajan - 3 years, 8 months ago

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@Ameya Daigavane This is basically the only relatively clean proof!

Sualeh Asif - 3 years, 8 months ago

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

Nihar Mahajan - 3 years, 8 months ago

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@MS HT @Sharky Kesa @ZK LIn

Abdur Rehman Zahid - 3 years, 8 months ago

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You really want my long, expanded solution?

Sharky Kesa - 3 years, 8 months ago

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Yes :P

Nihar Mahajan - 3 years, 8 months ago

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Yes please.

Abdur Rehman Zahid - 3 years, 8 months ago

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

Abdur Rehman Zahid - 3 years, 6 months ago

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@Abdur Rehman Zahid 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)

Sualeh Asif - 3 years, 6 months ago

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@Gurīdo Cuong

Laurent Michael - 3 years, 8 months ago

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