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Prove that if $ a,b,c $ are non-negative real numbers such that $a+b+c = 3$, then $$ abc(a^2 + b^2 + c^2) \le 3 $$

My attempt : I tried AM-GM inequality, tried to convert it to $a+b+c$, but I think I cannot get the manipulation of $abc$.

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Note that $(ab+bc+ac)^2 \ge 3abc(a+b+c) = 9abc$

Since, $(ab+bc+ac)^2 = \sum\limits_{cyc} a^2b^2 + 2\sum\limits_{cyc}ab^2c \ge 3abc(a+b+c)$

Hence, $$\begin{align}abc(a^2 + b^2 + c^2) &\le \frac{1}{9}(ab+bc+ac)^2(a^2 + b^2 + c^2) \tag{1} \\&\le \frac{1}{9}\left(\frac{a^2 + b^2 + c^2+2ab+2bc+2ac}{3}\right)^3\tag{2} \\ &= \frac{(a+b+c)^6}{3^5} = 3\tag{3}\end{align}$$

$(1)$ application of Am-Gm Inequality with three terms $$a^2+b^2+c^2,ab+bc+ca,ab+bc+ca$$

$(2)$ Using $a^2+b^2+c^2+2(ab+bc+ca) = (a+b+c)^2$

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  • $\begingroup$ can you explain the 2nd last step plz. @r9m $\endgroup$ – novak Jan 28 '15 at 21:15
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    $\begingroup$ It seems AM-GM for $a^2+b^2+c^2,\ ab+bc+ac,\ ab+bc+ac$. $\endgroup$ – Berci Jan 28 '15 at 21:17
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Let: $$ M_p = \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p} $$ for $p>0$ and $M_0 = \sqrt[3]{abc}$. Since $M_p$ is a log-concave and increasing function, $$ M_0^3 M_2^2 \leq M_{\frac{4}{5}}^5 \leq M_1^{5} = 1 $$ and the claim follows.

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