Let $a$, $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2=3$. Prove that: $$a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$$ I tried Rearrangement, uvw, BW and more, but without any success.
If $a^2+b^2+c^2=3$ so $a^2b+b^2c+c^2a\geq3\sqrt[3]{a^2b^2c^2}$
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inequality
contest-math
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0why should this help? – 2017-01-21
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0@MichaelRozenberg Have you considered bounding the $3\sqrt[3]{a^2b^2c^2}$ term w.r.t. the $a^2 + b^2 + c^2 = 3$ condition and then squeezing with a slightly stronger inequality? – 2017-01-28
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0@ÍgjøgnumMeg I tried. This inequality is very strong and I think we must use here strong methods. uvw for example. I lost here. – 2017-01-28
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0@MichaelRozenberg If we think about a cube with side lengths $x = a^2, y = b^2$ and $z = c^2$ we can bound the volume of the cube under the plane $x + y + z - 3 = 0$ via lagrange multipliers and then prove a slightly different inequality, would you agree? – 2017-01-28
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0@ÍgjøgnumMeg This inequality is cyclic and not symmetric. I think, Lagrange multipliers method does not help here. It's impossible to solve the system, which we get. I think. – 2017-01-28
1 Answers
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$$\Leftrightarrow (a^2b+b^2c+c^2a)^2 \ge 9(abc)^{4/3}=3(abc)^{4/3}(a^2+b^2+c^2)$$
$$\sum_{cyc}a^4b^2+2\sum_{cyc}a^3bc^2 \ge 3\sum_{cyc}\sqrt[3]{a^{10}b^4c^4}$$
By AM-GM : $$a^4b^2+2a^3bc^2 \ge 3\sqrt[3]{a^{10}b^4c^4}$$