How to solve $(x-2)(x^2+1)^{1/2}>(x^2+2)$? I am very much tempted to square both sides to obtain a solution, but I'm not sure whether doing so is a valid step. What conditions should I impose on $x$ in order to square both sides? 
In fact, this question leads me to the following problem: Under what conditions will $f(x)>g(x)$ imply $[f(x)]^2>[g(x)]^2$?
 A: To square and preserve the inequality, you need to be sure that both sides are greater than zero (where $f(x)=x^2$ is increasing). 
This requires $x-2> 0\Rightarrow x> 2$. It should be clear the other terms are positive, and you do not need to worry about them. Good question!
A: Note that $x^2+2>0$ and $\sqrt{x^2+1}>0$. Thus, in order the inequality is satisfied, you also need $x-2\ge0$: if $x<2$, the inequality is surely not satisfied.
Thus the inequality becomes a system of inequalities:
$$
\begin{cases}
x-2\ge0 \\[4px]
(x-2)^2(x^2+1)>(x^2+2)^2
\end{cases}
$$
The second inequality can be treated as usual:
$$
x^4-4x^3+4x^2+x^2-4x+4>x^4+4x^2+4
$$
or
$$
4x^3-x^2+4x<0
$$
This can be factored as
$$
x(4x^2-x+4)<0
$$
The quadratic factor has negative discriminant, so it's positive for every value of $x$. Hence the solution set is $x<0$, which, combined with $x\ge2$ tells us the inequality has no solution.

It would be very different if the inequality had been
$$
(x-2)\sqrt{x^2+1}<x^2+2
$$
Here, any value $x<2$ is a solution, because the left-hand side is negative and the right-hand side is positive.
For $x\ge2$, we can square and get
$$
\begin{cases}
x\ge2 \\[4px]
x(4x^2-x+4)>0
\end{cases}
$$
which has $x\ge2$ as solution set. Thus the inequality is satisfied for all values of $x$.
We already knew it, of course, but I wanted to emphasize the fact that just squaring would be wrong, because it would lead to $x(4x^2-x+4)>0$ and to the wrong conclusion that the inequality is only satisfied for $x>0$.
A: it must be $$x>2$$ and we can square both sides and we get
$$(x-2)^2(x^2+1)>x^2+2$$ this is equivalent to
$$-4x^3+x^2-4x>0$$
can you proceed?
after factoring we have
$$x(-4x^2+x-4)>0$$ since we have $$x>2$$ you have to solve
$$-4x^2+x-4>0$$
A: The inequality is never verified.
In fact, for $x\le 2$ it is obvious since $x^2+2\gt 0$. Equalizing $$(x-2)\sqrt{x^2+1}=x^2+1\iff 4x^3-x^2+4x=0$$ and there is just one real root $x=0$ so never there is equality for $x\gt 2$ Since for $x=3$ one has LHS is not greater than RHS (one would have the absurde $\sqrt{10}\gt 10$), we finish.
