KVPY SX SB Exam 2016 Question Paper

(x, y, z) are real & ${\mathit{x}}^4,{\mathit{y}}^4,{\mathit{z}}^4$ are positive real numbers
$\therefore \frac{{\mathit{x}}^4+{\mathit{y}}^4+{\mathit{z}}^4+1}{4}\ge \left|\mathit{x}\mathit{y}\mathit{z}\right|$
$\Rightarrow \left(\mathit{x}\mathit{y}\mathit{z}\right)\ge |\mathit{x}\mathit{y}\mathit{z}|$
i.e., xyz > 0
So it holds equality
$\therefore {\mathit{x}}^4={\mathit{y}}^4={\mathit{z}}^4=1$ ; But xyz > 0
∴ (x, y, z) ∈ {(1, 1, 1), (1, –1, –1), (–1, 1, –1), (–1, –1, 1)}
So no. of triplets is 4.