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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:

a, b, c are positive reals. Show that $a^{3} + b^{3} + c^{3} + 3abc \geq ab(a + b) + bc(b + c) + ca(c + a)$ .

Solution

Solution:

The inequality is homogeneous, so we can take $a = 1$ and put $b = 1 + x$ , $c = 1 + y$ , where $x, y \geq 0$ . Then after some reduction the inequality is equivalent to $x^{3} + y^{3} + x^{2} + y^{2} - x^{2} - y - xy^{2} - xy \geq 0$ , or (after factorising $x^{3} + y^{3}$ ) to $(x + y + 1)(x - y)^{2} + xy \geq 0$ , which is obviously true.

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