Proving $\forall x\in \mathbb R$, if $x>0$ then $(x+\frac 1 x \ge 2)$

Question

Prove $\forall x\in \mathbb R$, if $x>0$ then $(x+\frac 1 x \ge 2)$

I think a proof by contradiction is the easiest in this case, so we have: $\forall x\in \mathbb R :x>0\wedge \neg(x+\frac 1 x \ge 2)\iff \forall x\in \mathbb R :x>0\wedge (x+\frac 1 x < 2)$

Take $x=100$ and we get an immediate contradiction $100+\frac 1 {100}< 2$

Did I use proof by contradiction correctly?

Is this statement $\forall x (p\to q)$ or $(\forall x (p)) \to q$?

Answer 1

Multiply each side of the inequality by $x$.

Since $x>0$, the direction of the inequality will not invert, so we get:

• It's sufficient to prove that $x>0 \implies x^2+1\geq2x$
• It's sufficient to prove that $x>0 \implies x^2-2x+1\geq0$
• It's sufficient to prove that $x>0 \implies (x-1)^2\geq0$

The last one is obviously true for every $x\in\mathbb{R}$, hence true for $x>0$.

Answer 2

The easiest proof is by noting that $\left(x+\dfrac{1}{x}-2\cdot\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}\right)=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2\ge0$.

Answer 3

If $x > 0$, then there is a number $y > 0$ such that $y^2 = x$, hence $$x + \frac{1}{x} = y^2 + \frac{1}{y^2} = y^2 - 2 + \frac{1}{y^2} + 2 = \left(y - \frac{1}{y}\right)^2 + 2 \ge 0 + 2 = 2.$$ The inequality step is true because no real square is negative. Equality is therefore attained when $y = \frac{1}{y}$, or $y = 1$; i.e., $x = 1$.

Answer 4

By AM–GM inequality we have

$$ \frac{x+\dfrac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}}=1. $$

Generalisation: for $a_i >0$, such that $a_1 a_2 \ldots a_n =1$ we have $$ a_1+a_2+\cdots+a_n \geq n. $$