Question

For distinct positive numbers x, y, z, prove that (x +y)(y + z)(z + x) > 8xyz.

Answer 1

We have A > G

For positive numbers x and y, x + y > 2\(\sqrt{xy}\)

For positive numbers y and z, y + z > 2\(\sqrt{yz}\)

For positive numbers z and x, z + x > 2\(\sqrt{zx}\)

Multiplying, (x +y)(y + z)(z + x) > \(8\sqrt{x^2y^2z^2}\)

i.e., (x +y)(y + z)(z + x) > 8xyz