Prove: $ x + 9/x \geq 6$ for every positive real number $x$

Question

I’m at a loss as to how to prove this. I thought it would be a good idea to use a Direct Proof to tackle this problem, and thus solved it algebraically, but upon revising, the “for every positive real number x” confused me.

I revised my textbook and the recommendation was to try to prove this for some arbitrary real number c, but I’m not sure how to proceed here. What steps should I take when dealing with this type of problem?

Answer 1

Rewriting the inequality

$$x^2-6x+9\geq 0$$

$$(x-3)^2\geq 0$$

Alternative:

AM-GM:

$$\frac{x+\frac{9}{x}}{2}\geq \sqrt{9}$$

Answer 2

$x>0$;

$(\sqrt{x}-\sqrt{(9/x)})^2 \ge 0;$

$x +9/x -2 \sqrt{ x\cdot (9/x)} \ge 0$.