The Minimum Value

Let $x^{2}+\frac{1}{x^{2}}=y$

Differentiating on both sides, ${y}'= 2x-\frac{2}{x^{3}}$

$\therefore {y}''=2+\frac{6}{x^{4}}$

Now, to find the critical points of y,

${y}'= 2x-\frac{2}{x^{3}}=0$

$\therefore x^{4}-1=0$

$\therefore x^{2}=\pm1$

Now, $x^{2}=-1$ means that $x$ is a complex no.

Assuming that $x$ belongs to real numbers, I've discarded that possibility.

$\therefore x^{2}=1$ Now, $\therefore {y}''_{x^{2}=1}=2+\frac{6}{1} =8 \;\;\Rightarrow (+)ve$

$\therefore y_{x^{2}=1} \Rightarrow minimum\;value\;of\;y$

$\therefore y_{min}=y_{x^{2}=1}=1+\frac{1}{1}=2$

First use the fact that... (x - 1/x ) ^{2} > or = 0 simplify this.... x^2 - 2 + 1/x^2 > or = 0 by Addition property of inequality... x^2 + 1/x^2 > or = 2 the answer. However...... What if I used the Fact that... (x + 1/x)^2 > or = 0 so... x^2 +2 + 1/x^2 > or = 0 and this implies that... x^2 + 1/x^2 > or = -2 [ Prove that this is wrong ].