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# NMTC Problem

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

Given that $$x,y$$ and $$z$$ are positive reals, prove the inequality above.

Note by Abdur Rehman Zahid
1 year ago

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Here 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 · 1 year ago

Well, proving the Vasc Inequality might be slightly more difficult. · 1 year ago

To be exact Tremendously difficult

It does not yeild to most classical inequalities! · 1 year ago

Yes, I know. I doubt it can be quoted in contests either. · 1 year ago

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

This is basically the only relatively clean proof! · 1 year ago

Awesome!!! (speechless) · 1 year ago

Yep! Titu and Vasc killed this... =D · 1 year ago

@Gurīdo Cuong · 1 year ago

You really want my long, expanded solution? · 1 year ago

I still haven't seen your long and expanded solution Sharky :P · 11 months 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) · 11 months ago

Yes please. · 1 year ago