Show that $\forall x>0,\;\frac{1}{x}+x\geq 2$

$(x-1)^2\ge 0\iff x^2-2x+1\ge 0\iff x^2+1\ge 2x\iff x+\dfrac 1x\ge 2$ whenever $x>0$.

$f(x) = x + x^{-1}, \; \Bbb 0 < x \in \Bbb R; \tag 1$

$f'(x) = 1 - x^{-2} = 0 \Longrightarrow x = 1; \tag 2$

$f''(x) = 2x^{-3}; \tag 3$

$f''(1) = 2 \Longrightarrow 1 \; \text{is a local minimum of} \; f(x); \tag 4$

note that

$0 < x < 1 \Longrightarrow f'(x) < 0; \; 1 < x \Longrightarrow f'(x) > 0, \tag 5$

which implies that in fact $x = 1$ is a global minimum for $f(x)$; also,

$f(1) = 2, \tag 6$

and thus we conclude that

$f(x) \ge 2, \; \forall 0 < x \in \Bbb R. \tag 7$

$$ F(x)= x+ \frac 1x -2 , x>0$$ We can write, $$ F(x)=(\sqrt x)^2 + (\frac{1}{\sqrt x})^2 - 2.\sqrt x.\frac{1}{\sqrt x},{\ }since {\ }x>0$$
$$F(x)= (\sqrt x -\frac{1}{\sqrt x})^2$$ Thus we can conclude, $F(x) \geq 0 \forall x > 0$

Equality holds for x = 1.