<s>[INST] 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!
 [/INST] 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.
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