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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!

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    $\begingroup$ $a>b$ does not necessarily mean that $a^2>b^2$ (your second step) $\endgroup$ – Kenny Lau Jun 28 '16 at 16:38
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    $\begingroup$ @Kenny Lau I really didn't know that... Thanks for telling me! $\endgroup$ – T4yl0r Jun 28 '16 at 16:39
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    $\begingroup$ Indeed, $1 > -2$, but $1^2 < (-2)^2$ $\endgroup$ – G Tony Jacobs Jun 28 '16 at 16:43
  • $\begingroup$ Though the above comments are correct, the calculation mistake in your answer is in the values of $x_1$ and $x_2$. From $2x^2-4=0$, you get $x=\pm\sqrt2$ $\endgroup$ – GoodDeeds Jun 28 '16 at 16:45
  • $\begingroup$ Please note: Whatever you do, you will need an argument on how to turn the implications around, $\endgroup$ – AD. Jun 28 '16 at 16:49
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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 $$0<x<\sqrt2$$

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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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