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Let $a$, $b$ and $c$ be real numbers. Prove that: $$4(a^6+b^6+c^6)+5(a^5b+b^5c+c^5a)\geq\frac{(a+b+c)^6}{27}$$ I tried SOS, uvw and more, but without success.

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The following proof is true if they are non-negative reals.

$4(x^3+y^3+z^3)+15xyz\geq (x+y+z)^3$- this directly follows from Schur. So $$4(a^6+b^6+b^6)+5(a^5b+b^5c+c^5a)\geq 4(a^6+b^6+c^6)+15a^2b^2c^2\geq (a^2+b^2+c^2)^3\geq\dfrac{(a+b+c)^6}{27}$$

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  • $\begingroup$ Nice. But can you explain the first inequality ? How to prove $a^5b + b^5c +c^5a \geq 3a^2b^2c^2$ ? Thanks. $\endgroup$ – user230452 Dec 18 '16 at 6:39
  • $\begingroup$ it is AM-GM: $x+y+z\geq 3\sqrt[3]{xyz}$ $\endgroup$ – dezdichado Dec 18 '16 at 6:49
  • $\begingroup$ @dezdichado Your inequality $x+y+z\geq3\sqrt[3]{xyz}$ and your solution are wrong! Try $x=y=0$ and $z=-1$. $\endgroup$ – Michael Rozenberg Dec 18 '16 at 11:31
  • $\begingroup$ I see, they can be negative. $\endgroup$ – dezdichado Dec 18 '16 at 16:42
  • $\begingroup$ @dezdichado OMG. You're right, it's so obvious. I missed it. $\endgroup$ – user230452 Dec 19 '16 at 4:26
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With computer, here is a solution.

We have $$4(a^6+b^6+c^6) + 5(a^5b+b^5c+c^5a) - \frac{(a+b+c)^6}{27} = \frac{1}{324}z^TQz$$ where \begin{align} z = \left(\begin{array}{l} a^2\, b - a^3\\ a^2\, c - a^3\\ a^3 - 2\, a^2\, b + a\, b^2\\ a^3 - a^2\, c - a^2\, b + a\, b\, c\\ a^3 - 2\, a^2\, c + a\, c^2\\ - a^3 + 3\, a^2\, b - 3\, a\, b^2 + b^3\\ - a^3 + 2\, a^2\, b + c\, a^2 - a\, b^2 - 2\, c\, a\, b + c\, b^2\\ - a^3 + 2\, a^2\, c + b\, a^2 - a\, c^2 - 2\, b\, a\, c + b\, c^2\\ - a^3 + 3\, a^2\, c - 3\, a\, c^2 + c^3 \end{array}\right) \end{align} and \begin{align} Q = \left(\begin{array}{rrrrrrrrr} 21060 & -10530 & 17820 & 4515 & -8817 & 3685 & 4023 & -1722 & -1575\\ -10530 & 21060 & -6135 & -903 & 17820 & -882 & -2166 & 1785 & 3783\\ 17820 & -6135 & 18550 & 1719 & -4026 & 4590 & 3570 & -2286 & -285\\ 4515 & -903 & 1719 & 6108 & -3450 & -60 & 1485 & 1320 & -1332\\ -8817 & 17820 & -4026 & -3450 & 18354 & -279 & -2115 & 792 & 4590\\ 3685 & -882 & 4590 & -60 & -279 & 1284 & 774 & -642 & 126\\ 4023 & -2166 & 3570 & 1485 & -2115 & 774 & 1104 & -246 & -485\\ -1722 & 1785 & -2286 & 1320 & 792 & -642 & -246 & 790 & -36\\ -1575 & 3783 & -285 & -1332 & 4590 & 126 & -485 & -36 & 1284 \end{array}\right). \end{align} It is easy to verify that $Q$ is (real symmetric) positive definite. We are done.

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