# 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.

• $f(x)=x+1/x\implies f(x)=f(1/x)\implies f(1)=2$ is a relative maxima or minima of $f(x)$, since $f(x)=f(1/x)$, so no other points can be relative maxima or minima since every other point is equal to another, then it is nearly needed to graph to see whether it is a maxima or minima. Aug 9, 2016 at 15:23

With AM-GM we get $$x+\frac { 1 }{ x } \ge 2\sqrt { x\frac { 1 }{ x } } =2$$

• (which is obviously attained by a specific value of $x$) Aug 9, 2016 at 15:20
• "with equality if and only if $x = \frac{1}{x}$''. Aug 9, 2016 at 15:36

Hint. One may observe that, $$\left(x+\frac1x \right)^2-4=\left(x-\frac1x \right)^2\ge0, \qquad x>0.$$

• This turns out to be exactly the same as the other answer, since this consideration is one way of proving the AM-GM inequality. Aug 9, 2016 at 15:23
• @MorganRogers Actually, this answer uses considerably weaker assumptions than AM-GM. The AM-GM can be proved from the principle that no (real) square is negative, but it is a stronger result, consequently stronger than is needed to prove the original statement. Aug 9, 2016 at 15:27
• @heropup For AM-GM the only assumption is that the terms are real and non-negative, so I'm pretty sure it's the same..? Aug 9, 2016 at 15:34
• @MorganRogers $$(x_1 - x_2)^2 \ge 0$$ is definitely a weaker claim than $$(x_1 + \cdots + x_n)/n \ge (x_1 \cdot \ldots \cdot x_n)^{1/n}.$$ To prove the latter from the former, one would probably apply some induction argument. Aug 9, 2016 at 15:40
• @heropup You can indeed prove the latter from the former by induction, with no further assumptions. Then given the latter at $n=2$, $$(x_1+x_2)/2 \geq (x_1 x_2)^{1/2} \implies (x_1 - x_2)^2 \geq 0.$$ So the two are equivalent. Aug 10, 2016 at 11:45

Let $f(x)=x+\frac{1}{x}$. Clearly $f(x)=f(1/x)$, so we minimize in $[1,\infty)$

Notice $x+\frac{1}{x}>1+1\iff x-1>1-\frac{1}{x}=\frac{x-1}{x}$, which is clear since $x\geq 1$

$x+1/x\ge m$ for all positive $x$ iff the same $x$ satisfy $x^2-mx+1\ge0$. Clearly $m\le2$. But if $m<2$ then $x^2-mx+1>(x-1)^2\ge0$, so $m=2$.

$(x-1)^2 \geq 0$ for $x>0$. This, in turn, implies that $x+1/x\geq 2$ for all $x>0$. I have left the details. So, the min value of $x+1/x$ is 2 on $x>0$.