Find the min value of $x+1/x$ on $(0, \infty)$ without using calculus. Using calculus, it is pretty obvious. I was trying to see if it can be done without calculus. May be some sort of inequalities.
With AM-GM we get $$x+\frac { 1 }{ x } \ge 2\sqrt { x\frac { 1 }{ x }  } =2$$
$(x-1)^2 \geq 0$ for $x&gt;0$. This, in turn, implies that $x+1/x\geq 2$ for all $x&gt;0$. I have left the details. So, the min value of $x+1/x$ is 2 on $x&gt;0$.