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?

Have you heard of arithmetic mean -geometric mean inequality? — Akash Yadav, Mar 29 '20 at 16:13
You could also find by calculus that the minimum of $x + 9/x$ for $x>0$ occurs at $x=3$. — Ned, Mar 29 '20 at 18:42

score 2 · Answer 1 · answered Mar 29 '20 at 16:13

2

Rewriting the inequality

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

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

Alternative:

AM-GM:

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

answered Mar 29 '20 at 16:13

h-squared

1,333
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score 0 · Answer 2 · answered Mar 29 '20 at 16:24

0

$x>0$;

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

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

answered Mar 29 '20 at 16:24

Peter Szilas

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

2 Answers2