<s>[INST] Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt{x^2-10x+9}\left(2x+1\right)\ge \:\sqrt{x^2-10x+9}$$ Therefore $ 2x+1 \ge 0 => x\ge \frac{1}{2} $ so $x$ belongs to $(\frac{1}{2},\infty) $ and $ x^2-10x+9 \ge 0 => x$ belongs to $ (-\infty,1] $ and $[9,\infty) $. Therefore the statements are contradicting each other.
 [/INST] Hint:
The domain of the inequation is defined by
$$x^2-10x+9\ge 0\iff (x-5)^2\ge 16\iff x\ge 9\quad\text{or}\quad x\le 1.$$
$u\ge \sqrt u\iff u\ge 1$ or $u=0$. Here 
$$u=0 \iff\sqrt{x^2-10x+9}=-x\iff -10x+9=0\quad\text{and}\quad x\le 0,$$
which has no solution. Hence the inequation comes down to
$$\sqrt{x^2-10x+9}\ge 1-x\iff x^2-10x+9\ge (1-x)^2\quad\textbf{or}\quad x\ge 1. $$
Can you take it from here?
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