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Let $a, b, c$ be positive real numbers. Show that
$$a^{3}+b^{3}+c^{3} \geq a^{2} b+b^{2} c+c^{2} a .$$
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We have
$$\begin{array}{r} a^{3}+a^{3}+b^{3} \geq 3 \sqrt[3]{a^{3} a^{3} b^{3}}=3 a^{2} b \\ b^{3}+b^{3}+c^{3} \geq 3 \sqrt[3]{b^{3} b^{3} c^{3}}=3 b^{2} c \\ c^{3}+c^{3}+a^{3} \geq 3 \sqrt[3]{c^{3} c^{3} a^{3}}=3 c^{2} a \end{array}$$
Adding up these inequalities, we get that
$$3 a^{3}+3 b^{3}+3 c^{3} \geq 3 a^{2} b+3 b^{2} c+3 c^{2} a$$
Hence, the result follows when we divide throughout by $3 .$
by Diamond (75,914 points)

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