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.
 A: 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 $y\ge{8}$. 


*

*$y\ge{8}$ 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.


*

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

A: 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?
A: Using Kenny Lau's substitution above: 
$u=x+\sqrt{x^2-10x+9}$ gives us that $u\geq\sqrt{u}$ and hence $u\geq1$. (Do you see why this last part is true? Don't move on until you do!)
So, $x+\sqrt{x^2-10x+9}\geq 1$. Now, rearraning to get the root on its own, we have $x-1\geq-\sqrt{x^2-10x+9}$. Here comes the tricky part. The right side is clearly negative, but what about the left? We just don't know, truthfully. If $x=0$ it is negative, but if $x=2$ it is positive, for example. 
We can't just square both sides. We need to be SURE we have a negative on both sides or a positive on both sides to do that. But there is another way.
What are the roots of the expression under the radical? Factoring gives $x^2-10x+9=(x-1)(x-9)$, which has roots $x=1,9$. Try plugging in some values in the intervals $(-\infty,1), (1,9)$, and $(9,\infty)$.
You will see that for $(1,9)$, the root has a negative argument, which makes no sense here, so $x$ does not lie in that range. But otherwise, it has a positive argument. On $(9,\infty)$, it is obvious to me that the inequality is satisfied (but is it obvious to you? Try some values!). On $(-\infty,1)$, it is a little more subtle. Since you have asked other questions relating to calculus, it is a nice exercise to show that for $x\leq1$, the function $f(x)=x+\sqrt{x^2-10x+9}$ is greater than $1$ (which is what you wanted!). I will leave that bit to you. :)
We have equality in our problem, hence we have the following solution: $x\in(-\infty,1]\cup[9,\infty)$.
A: The suggestion about $u\ge\sqrt{u}$ helps in removing one level of square roots. Now you have
$$
x+\sqrt{x^2-10x+9}\ge1
$$
or
$$
\sqrt{x^2-10x+9}\ge1-x
$$
If you notice that the expression under the square root is $(x-1)(x-9)$, you can possibly use the clever trick shown in another answer. On the other hand, such an inequality can be analyzed in a systematic way, by transforming it into two systems of inequalities:
$$
\mathrm{(I)}
\begin{cases}
x^2-10x+9\ge0 \\[4px]
1-x<0
\end{cases}
\qquad\text{or}\qquad
\mathrm{(II)}
\begin{cases}
\bigl[\,x^2-10x+9\ge0\,\bigr] \\[4px]
x^2-10x+9\ge(1-x)^2 \\[4px]
1-x\ge0
\end{cases}
$$
The first system comes from the consideration that the square root is non-negative as soon as it exists, so it will be greater than a negative number; the second system comes from the consideration that an inequality between non-negative numbers is equivalent to the same inequality between their squares.
System (I) can be transformed into
$$
\mathrm{(I)}
\begin{cases}
x\le1 \quad\text{or}\quad x\ge9 \\[4px]
x>1
\end{cases}
$$
which has $x\ge9$ as solution set.
System (II), where the top inequality is implied by the middle one, becomes
$$
\mathrm{(II)}
\begin{cases}
x^2-10x+9\ge1-2x+x^2 \\[4px]
x\le1
\end{cases}
$$
and it is easily seen that the solution set is $x\le1$.
Summing up, the solution set of the original inequality is
$$
\boxed{\quad\vphantom{\Big|}x\le 1\quad\text{or}\quad x\ge9\quad}
$$
