<s>[INST] 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 $1<x<3$?
 [/INST] 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) $x+9/x \ge 6.$
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