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Algebra Difficulty 7.4 National olympiad, round 2 Prove it

Example 6 Let a,b,c∈R+a, b, c \in \mathbf{R}^{+}, prove:
a+b+c3⩾(a+b)(b+c)(c+a)83⩾ab+bc+ca3\begin{array}{c} \frac{a+b+c}{3} \geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}} \\ \geqslant \frac{\sqrt{a b}+\sqrt{b c}+\sqrt{c a}}{3} \end{array}

Solution

Prove that by the AM-GM inequality, we have
(a+b)(b+c)(c+a)3⩾(a+b)(b+c)(c+a)3\begin{aligned} & \frac{(a+b)(b+c)(c+a)}{3} \\ \geqslant & \sqrt[3]{(a+b)(b+c)(c+a)} \end{aligned}

Therefore, a+b+c3\frac{a+b+c}{3}
⩾(a+b)(b+c)(c+a)83\geqslant \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}

By (7) (Hölder's inequality), we have
(a+b)(b+c)(c+a) 8 = 1 27 ( a+b 2 +b+a ) (b+ b+c 2 +c ) (a+c+ a+c 2 ) 1 27 ( [3] a+b 2 b a + [3] b b+c 2 c . .+ [3] a c a+c 2 ) 3 1 27 ( [3] a b a b + [3] b c b c + [3] c a c a ) 3 = 1 27 ( a b + b c + c a ) 3\text{(a+b)(b+c)(c+a) 8 = 1 27 ( a+b 2 +b+a ) (b+ b+c 2 +c ) (a+c+ a+c 2 ) 1 27 ( [3] a+b 2 b a + [3] b b+c 2 c . .+ [3] a c a+c 2 ) 3 1 27 ( [3] a b a b + [3] b c b c + [3] c a c a ) 3 = 1 27 ( a b + b c + c a ) 3}

Therefore, (a+b)(b+c)(c+a)83\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}
⩾ab+bc+ca3\geqslant \frac{\sqrt{a b}+\sqrt{b c}+\sqrt{c a}}{3}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.