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

Question

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 $1<x<3$?

Answer 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.

Answer 2

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 $x+\frac9x\geqslant6$.

Answer 3

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.

Answer 4

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