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Algebra Difficulty 5.9 AIME, harder Prove it

Example 2.2.7. Let a,b,ca, b, c be positive real numbers. Prove that
a2(b+c)+b2(c+a)+c2(a+b)≥(ab+bc+ca)(a+b)(b+c)(c+a)3a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geq(a b+b c+c a) \sqrt[3]{(a+b)(b+c)(c+a)}

Solution

Solution. Notice that the following expressions are equal to each other
a2(b+c)+b2(c+a)+c2(a+b)b2(c+a)+c2(a+b)+a2(b+c)ab(a+b)+bc(b+c)+ca(c+a)\begin{array}{l} a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\ b^{2}(c+a)+c^{2}(a+b)+a^{2}(b+c) \\ a b(a+b)+b c(b+c)+c a(c+a) \end{array}

According to Hölder's inequality, we get that
(∑cyca2(b+c))3≥(∑cycab(a+b)(b+c)(c+a)3)3\left(\sum_{c y c} a^{2}(b+c)\right)^{3} \geq\left(\sum_{c y c} a b \sqrt[3]{(a+b)(b+c)(c+a)}\right)^{3}
which is exactly the desired result. Equality holds for a=b=ca=b=c.

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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.