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 $1<x<3$?
 A: 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.
A: 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 $x+\frac9x\geqslant6$.
A: 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.
A: Option:
AM-GM:
$x>0.$
$x+9/x \ge 2\sqrt{x(9/x)}= 2\sqrt{9} =6.$
Equality for $x=9/x$.
Your answer, as other users have pointed out already:
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) $x+9/x \ge 6.$
