<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] You have got yourself into a tangle by being too clever too soon, and squaring. It is better to simplify first. With careful simplification you don't need calculus or anything advanced at all.
To pick up a theme of Kenny Lau, put $u=x+\sqrt{x^2-10x+9}$. Then your inequality becomes $u\ge\sqrt{u}$, which is equivalent (following Gerry Myerson's comment) to $u\ge{1}$.
Translating back to $x$ language, this makes your inequality equivalent to $$x+\sqrt{x^2-10x+9}\ge{1}$$
Since the expression under the square root is equivalent to $(x-1)(x-9)$, we can write this as $$(x-1)+\sqrt{(x-1)(x-9)}\ge{0}$$ and putting $y=x-1$, we get $$y+\sqrt{y(y-8)}\ge{0}$$.
For the expression under the square root to make sense we need either $y\le{0}$ or $y\ge{8}$. 


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*$y\ge{8}$ certainly satisfies the inequality. Put into $x$ terms, this means $x\ge{9}$.


What about $y\le{0}$? Let's do another substitution, of $z=-y$, which means that $z\ge{0}$. This gives us $$-z+\sqrt{z(z+8)}\ge{0}$$
Dividing by $\sqrt{z}$ and rearranging, we get $$\sqrt{z+8}\ge\sqrt{z}$$
… which is always true.


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*Therefore $z\ge{0}$ also satisfies the equality, which means $y\le{0}$, which means $x\le{1}$.

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