# Prove the inequality $x+\frac{1}{x}\geq 2$? [duplicate]

I'm reading Courant's Calculus. There is an exercise: Prove the inequality $x+\frac{1}{x}\geq 2$.

I did the following: For the sake of curiosity, I did:

$$x+\frac{1}{x}\geq 2$$

$$x^2+1\geq 2x$$

$$x^2-2x +1\geq 0$$

And this is just $(x-1)^2$. Well, as the square of a real number is $\geq0$, then:

$$(x-1)^2 \geq 0$$

$$x^2-2x+1 \geq 0$$

$$x^2+1 \geq 2x$$

$$x+\frac{1}{x} \geq 2$$

Which is what was to be proved. Is it correct?

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• That looks fine; You should probably justify dividing by $x$, which is okay when $x>0$. – Julius Aug 1 '16 at 5:54
• It is customary to omit the "scratch work" (first half). The second half is almost completely good. There is the unnecessary line $x^2+1\ge 0$. But the biggest flaw is the total absence of mention of the condition $x\gt 0$. Note that the inequality is false for $x\lt 0$. Some connecting words would be nice. – André Nicolas Aug 1 '16 at 5:59
• You have missed the restriction $x>0$, without it your multiplying through by $x$ does not necessarily preserve the inequality. – Conrad Turner Aug 1 '16 at 6:03
• @AndréNicolas The $x^2+1\ge 0$ was a typo. I noticed it now. – Billy Rubina Aug 1 '16 at 6:19
• You "may not" multiply by $x$ without discussing the sign. – Yves Daoust Aug 1 '16 at 6:32

By AM-GM, for all $x>0$:

$$\frac{x+\frac{1}{x}}{2} \geq\sqrt{x\cdot\frac{1}{x}}$$ $$x+\frac{1}{x} \geq2$$

You can write $$(x-1)^2 \ge 0 \\ x^2 - 2x + 1 \ge 0 \\ x^2 + 1 \ge 2x \\ x + \frac{1}{x} \ge 2,$$ omitting the third line. However, in the last step of the above--namely, division by $x$--you should note that this only preserves the direction of inequality if $x > 0$. Otherwise, we would reach the conclusion $$x + \frac{1}{x} \le -2, \quad x < 0.$$ Therefore, the original inequality to be shown is true only if $x > 0$.

$$x+ \frac 1 x = \left( x - 2 + \frac 1 x \right) + 2 = \underbrace{\left( \sqrt x - \frac 1 {\sqrt x} \right)^2} {}+ 2$$ The part over the $\underbrace{\text{underbrace}}$ cannot be negative, since it is a square, and is $0$ only when $x=1$.

• Ok, but this assumes $x\ge0$ upfront. – Yves Daoust Aug 1 '16 at 6:30

In all rigor, you can transform

$$x+\frac1x\ge2$$ in $$\frac{x^2-2x+1}x=\frac{(x-1)^2}x\ge 0.$$

Then the inequality only holds when the numerator and denominator have the same sign, which requires

$$x>0.$$

By the way, the LHS is an odd function, so that if for some $x$ you fulfill $x+\dfrac1x\ge2$, then $(-x)+\dfrac1{(-x)}\le-2$. So the inequality cannot hold for all $x$.

Assume $x\gt0,$ otherwise the inequality is wrong.

Either $x\ge1\ge\frac1x$ or else $x\le1\le\frac1x,$ i.e., $x-1$ and $1-\frac1x$ have the same sign,
whence $$(x-1)(1-\frac1x)\ge0$$ and $$x+\frac1x=2+(x-1)(1-\frac1x)\ge2.$$