TITLE: Show that $\forall x>0,\;\frac{1}{x}+x\geq 2$
QUESTION [7 upvotes]: Using optimality conditions, we want to show that for all positive $x$, we have $$\frac{1}{x}+x\geq 2\quad (1).$$
I think that the above problem is equivalent to showing that $1/x+x-2\geq 0, \forall x>0$. Let $f(x)=1/x+x-2$. Then I used the optimality conditions to get: $$f'(x)=-\frac{1}{x^2}+1$$ which is equal to zero iff $x=1$. And also $$f''(x)=-\frac{2}{x^3}$$ which is positive iff $x$ is negative or negative iff $x$ is positive. So, for $x>0$ we have a maximum point at $x=1$. But can we show that $f$ is non-negative?
I am not sure what I did wrong here. I'd appreciate any help. Thank you.

REPLY [4 votes]: $(x-1)^2\ge 0\iff x^2-2x+1\ge 0\iff x^2+1\ge 2x\iff x+\dfrac 1x\ge 2$ whenever $x>0$.