Discussion

Explanation:

Given, x + y + z = 6.

Let us consider three numbers (x + y), (y + z) and (z + x)

We know, AM ≥ GM

⇒ $\frac{(x+y)+(y+z)+(z+x)}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ $\frac{2(x+y+z)}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ $\frac{2\times 6}{3}$ ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ 4 ≥ $\sqrt[3]{(x + y)(y + z)(z + x)}$

⇒ (x + y)(y + z)(z + x) ≤ 64

Hence, 64.

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