What's the mistake on my answer for this inequality $ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $ Good evening to everyone! I have the following inequality: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $$. I don't know what's wrong with my answer: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} \rightarrow \left(x+1\right)\sqrt{x^2+4}>\left(x+2\right)\sqrt{x^2+1} \rightarrow \left(x+1\right)^2\left(x^2+4\right)>\left(x+2\right)^2\left(x^2+1\right) \rightarrow \left(x^2+2x+1\right)\left(x^2+4\right)>\left(x^2+4x+4\right)\left(x^2+1\right) \rightarrow 2x^3-4x<0 \rightarrow x\left(2x^2-4\right)<0 \rightarrow $$ $x$ belongs to $(-\infty,0)$ and $2x^2-4<0 \rightarrow x_1 = 1-\sqrt{2}$ and $x_2=1+\sqrt{2}$ therefore the final result is $x$ belongs to $(1-\sqrt{2},0) $. But my answer sheet shows that $x$ belongs to $(0,\sqrt{2})$. Where am I wrong? Thanks for any responses!
 A: ALthough in fact $a>b$ does not necessarily mean that $a^2 > b^2$ even if $a$ and $b$ are both positive, the real place you went wrong is in solving $2x^2-4 = 0$.  The answer to that is $x = \pm \sqrt{2}$ not $1 \pm \sqrt{2}$.
The $-\sqrt{2}$ zero is spurious; it was introduced by the quaring operation, and for $x=-\sqrt{2}$ you will find that the right hand side is positive and the left hand side negative (although the match in absolute value).  So the interesting interval is $(0,2)$ and it is easy to check, by trying $x=1$ that the left side minus the right side is positive in that interval.
A: The first part is correct. The inequality holds iff $$(x+1)\sqrt{x^2+4}>(x+2)\sqrt{x^2+1}\ \ (*)$$ Hence (squaring), provided $x+1>0$ (and hence $x+2>0$) we have $(x+1)^2(x^2+4)>(x+2)^2(x^2+1)$ which is equivalent to (expanding etc) $$2x(2-x^2)>0$$ which implies $0<x<\sqrt2$.
On the other hand if $-2<x<-1$ we have $x+1<0,x+2>0$ and so $(*)$ is false.
Whilst if $x<-2$, then $x+1<0,x+2<0$ and so $(*)$ is equivalent to $$|x+1|\sqrt{x^2+4}<|x+2|\sqrt{x^2+1}$$ Squaring as before this is equivalent to $$2x(2-x^2)<0$$ which this time is always false (for $x<-2$).
So, summarising, the given inequality is true iff $$0<x<\sqrt2$$
