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Algebra Difficulty 7.1 National olympiad, round 2 Prove it

28. Let a,b,ca, b, c be positive numbers, prove:
 (1) 4(a3+b3)⩾(a+b)3 (2) 9(a3+b3+c3)⩾(a+b+c)3\begin{array}{l} \text { (1) } 4\left(a^{3}+b^{3}\right) \geqslant(a+b)^{3} \\ \text { (2) } 9\left(a^{3}+b^{3}+c^{3}\right) \geqslant(a+b+c)^{3} \end{array}
(1996 British Mathematical Olympiad)

Solution

28. (1)−(1)-
a3+(−a+b2)3+(a+b2)3⩾3a(a+b2)2b3+(a+b2)3+(a+b2)3⩾3b(a+b2)2\begin{array}{l} a^{3}+\left(-\frac{a+b}{2}\right)^{3}+\left(\frac{a+b}{2}\right)^{3} \geqslant 3 a\left(\frac{a+b}{2}\right)^{2} \\ b^{3}+\left(\frac{a+b}{2}\right)^{3}+\left(\frac{a+b}{2}\right)^{3} \geqslant 3 b\left(\frac{a+b}{2}\right)^{2} \end{array}

Adding the two equations and simplifying yields the result.
(2)
a3+(a+b+c3)3+(a+b+c3)3⩾3a(−a+b+c3)2b3+(a+b+c3)3+(a+b+c3)3⩾3b(a+b+c3)2c3+(a+b+c3)3+(a+b+c3)3⩾3c(a+b+c3)2\begin{array}{l} a^{3}+\left(\frac{a+b+c}{3}\right)^{3}+\left(\frac{a+b+c}{3}\right)^{3} \geqslant 3 a\left(-\frac{a+b+c}{3}\right)^{2} \\ b^{3}+\left(\frac{a+b+c}{3}\right)^{3}+\left(\frac{a+b+c}{3}\right)^{3} \geqslant 3 b\left(\frac{a+b+c}{3}\right)^{2} \\ c^{3}+\left(\frac{a+b+c}{3}\right)^{3}+\left(\frac{a+b+c}{3}\right)^{3} \geqslant 3 c\left(\frac{a+b+c}{3}\right)^{2} \end{array}

Adding the three equations and simplifying yields the result.
Using this method, the general power mean inequality can be derived.

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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.