# Prove that x + $\frac{9}{x}$ $\ge$ 6 for all real numbers x $>$ 0

I have:

$$x^2+9\ge6x$$

$$x^2-6x+9\ge0$$

$$(x-3)^2\ge0$$

Is this a sufficient proof for all real numbers? Or do I need to prove that it works from $$1?

This is essentially right, but goes in the wrong order: you should start from the known true fact $$(x-3)^2\geq 0$$ and deduce what you want. This is sufficient for all positive real $$x$$. You use the fact that $$x>0$$ when you go from $$x^2+9\geq 6x$$ to $$x+9/x\geq 6$$; if $$1/x<0$$ the inequality would flip over at this point.

The logic in what you write is not quite clear to me. It seems you are assuming $$x^2+9 \geq 6x$$. Once you get the idea, which is indeed correct, it is better to write as follows: since $$(x-3)^2 \geq 0$$ for any real number $$x$$, then \begin{align} & x^2-6x+9 \geq 0 \\ & x^2+9 \geq 6x, \end{align} so when $$x > 0$$ we can divide by $$x$$ without changing the inequality and the claim is proved.

No, that does no prove it for all real numbers. First of all, you cannot possibly prove it for $$0$$, since what you wish to prove doesn't make sense in that case. On the other hand, you passed from $$x+\frac9x\geqslant6$$ to $$x^2+9\geqslant6x$$ multiplying by $$x$$. This can only be done if $$x>0$$; if $$x<0$$, the inequality must be reversed.

Note that $$x<0\implies x+\frac9x<0$$. Therefore, in that case you don't have $$x+\frac9x\geqslant6$$.

Option:

AM-GM:

$$x>0.$$

$$x+9/x \ge 2\sqrt{x(9/x)}= 2\sqrt{9} =6.$$

Equality for $$x=9/x$$.

For $$x>0$$.

Start from:

1)$$(x-3)^2 \ge 0$$, this is true for any real $$x$$ (why?)

Expand :

2) $$x^2-6x +9\ge 0.$$

Divide inequality by $$x>0:$$

3) $$x -6 +9/x \ge0.$$

4) $$x+9/x \ge 6.$$