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This is a math problem from the German Math Olympiad, but in this case I do not know where to start, probably because I do not have enough intuition regarding inequalities.

However, I tried to apply the standard AM-GM equation which didn't help, and in general I could not find a good up-/ downward estimation for the root term. I also looked at the function plots and it seems that those three terms are very close to each other which seems to make this problem even more difficult to prove.

Any tips on how to begin? Please do not post a complete solution here.

BTW usually those problems provide an elegant solution based on non-university level knowledge, so I do not like to use an approach like a series expansion of the root term!

EDIT: Alright, the right side should've been obvious: $$0<\frac{1}{4x^2}\Leftrightarrow \sqrt{x^2+1}<\sqrt{x^2+1+\frac{1}{4x^2}}=\sqrt{x^2+2x\frac{1}{2x}+\frac{1}{4x^2}}=x+\frac{1}{2x}$$

EDIT 2: So, I noted that squaring the left inequality (considering $x>0$ as well as $LHS>0$) gives me an inequality equivalent to $x\gt\sqrt{\frac{1}{8}}$. Therefore, I only have to show that the Assumption $LHS\gt 0$ is implying exactly this. I thus noted that $0=x+\frac{1}{2x}-\frac{1}{8x^3}$ has only one positive solution $\frac{\sqrt{\sqrt{3}-1}}{2}\gt\sqrt{\frac{1}{8}}$ -- but how do I show this as an inequality? I always get very confused about inequations with quadratic polynomials, especially regarding the direction of the inequality sign...

EDIT 3: Got it! So, let me state the whole proof ;)

To prove: $L:=x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}=:R,\quad x>0$

a) Consider $L\leq0$. $x>0$ implies $R>0 \Rightarrow L<R$

b) Consider $L\gt0$. We get $$ 0<x+\frac{1}{2x}-\frac{1}{8x^3} \Rightarrow 0\lt x^4+\frac{1}{2}x^2-\frac{1}{8} = (x^2)^2 + 2\frac{1}{4}x^2+\frac{1}{4^2}-\frac{1}{4^2}-\frac{1}{8} = (x^2+\frac{1}{4})^2-\frac{3}{16}\\ \Leftrightarrow \frac{\sqrt{3}}{4}\lt x^2+\frac{1}{4} \Leftrightarrow x^2 \gt \frac{\sqrt{3}-1}{4} \Leftrightarrow x \gt \sqrt{\frac{\sqrt{3}-1}{4}} \Rightarrow x\gt \sqrt{\frac{1}{8}} $$

From that, we get $$ \frac{1}{8x^2}\lt 1 \Leftrightarrow \frac{1}{8x^2}\frac{1}{8x^4}\lt \frac{1}{8x^4} \Leftrightarrow (\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt0 \\ \Leftrightarrow x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}\lt x^2+1 $$

Noticing that $L^2=x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}$ and $R^2=x^2+1$ gives you $$ \Leftrightarrow L^2 \lt R^2 \Leftrightarrow L\lt R $$ ...quod erat demonstrandum.

Sorry it took so long, I am not used to write that much TeX ;) BTW, can anyone leave a comment regarding the "beauty" of my proof? I tried to compose this final version as clean as possible!

Lukas Juhrich
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3 Answers3

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A start: Square everything. The inequality on the right will be obvious. The one on the left is a little more unpleasant.

André Nicolas
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  • You're right, that should've been obvious... brb trying the left side! – Lukas Juhrich Sep 10 '14 at 15:20
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    There is useful cancellation on the left when you square. You do have to play around a bit. Roughly speaking for the $x$ that could give trouble after the squaring, the expression on the left is less $1$ (and possibly negative, making the inequality clear). – André Nicolas Sep 10 '14 at 15:26
  • So I get $(x+\frac{1}{2x}-\frac{1}{8x^3})^2=x^2+1+(\frac{1}{8x^3})^2-\frac{1}{8x^4}$. Thus, I have to prove that $(\frac{1}{8x^3})^2-\frac{1}{8x^4}<0$, right? But I don't get where to get from there, the statement seems quite wrong... – Lukas Juhrich Sep 10 '14 at 15:57
  • hmm, just noted that the last inequality is equivalent to $x>\sqrt{\frac{1}{8}}$... can I construct this condition by the previous assumption that the LHS is greater than zero? – Lukas Juhrich Sep 10 '14 at 16:04
  • Got the concept, see my edit. – Lukas Juhrich Sep 10 '14 at 16:19
  • So you noticed that below $\sqrt{1/8}$ there is no trouble, because the left side is less than $1$. There is need to find roots, fairly crude estimates are good enough. – André Nicolas Sep 10 '14 at 16:41
  • Thanks, I could finish it :) hope my approach has been realized correctly – Lukas Juhrich Sep 10 '14 at 16:58
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    It is on the awkward side. Typo at the beginning of b), you left out the denominator. Too many fractions, should be able to show much faster that we are OK if $\frac{1}{x}\lt\sqrt{8}$. Then show that if $\frac{1}{x}\ge \sqrt{8}$ the original left side is less than $1$. Lots of slack! – André Nicolas Sep 10 '14 at 17:10
  • Where did I leave the denominator? The factor $\frac{1}{x^3}$ could be left because of $x>0\Rightarrow \frac{1}{x^3}>0$ – Lukas Juhrich Sep 10 '14 at 17:26
  • I called it a "typo." You wrote $x+\frac{1}{2x}-\frac{1}{8x^3}=\dots$. Of course you did not mean they are equal. – André Nicolas Sep 10 '14 at 17:31
  • Let us continue this discussion in chat. – Lukas Juhrich Sep 10 '14 at 17:36
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    After a couple of unpleasant experiences, I stopped using chat. When writing up contest questions, one has to be very careful. I have graded Olympiads both national and international (APMO). The grading is quite unforgiving. – André Nicolas Sep 10 '14 at 17:48
  • I do not need to care about the grading as I finished school this year. I use these problems as an exercise for myself. However it is true that I shouldn't state something which is not mathematically correct. What do you mean by "writing up" those questions? I kind of don't get the relation. – Lukas Juhrich Sep 10 '14 at 18:12
  • I was being imprecise, should have said contest solutions. One may have the right idea in a contest question, but if the write-up of the solution is not exact, the grade can drop precipitously from the standard $7$. In your solution, it would have been easier to square $x+\frac{1}{2x}-\frac{1}{8x^3}$ explicitly. The fact that we are OK for $\frac{1}{x}\le \sqrt{8}$ drops out immediately. – André Nicolas Sep 10 '14 at 19:07
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Just squaring everything seems the best way: $$ \left(x+\frac1{2x}\right)^2 =x^2+1+\frac1{4x^2} $$ and $$ \begin{align} \left(\color{#C00000}{x+\frac1{2x}}-\frac1{8x^3}\right)^2 &=\color{#C00000}{x^2+1+\frac1{4x^2}}-\frac2{8x^3}\left(\color{#C00000}{x+\frac1{2x}}\right)+\frac1{64x^6}\\ &=x^2+1-\frac1{8x^4}+\frac1{64x^6} \end{align} $$ Now you need to consider what happens to $x+\dfrac1{2x}-\dfrac1{8x^3}$ when $\dfrac1{64x^6}\gt\dfrac1{8x^4}$.

Now that the question seems answered, I might as well add the last little bit.

If $8x^2\lt1$, then $$ \begin{align} x+\frac1{2x}-\frac1{8x^3} &=\frac{8x^4+4x^2-1}{8x^3}\\ &\le\frac{x^2+4x^2-1}{8x^3}\\ &\le\frac{\frac58-1}{8x^3}\\ &=-\frac3{64x^3}\\[4pt] &\lt-\frac3{\sqrt8} \end{align} $$

robjohn
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It might be helpful to note that

$$\begin{align}\sqrt{x^2+1}&=x\sqrt{1+\frac 1{x^2}}\\ &=x\left(1+\frac 1{x^2}\right)^{\frac 12}\\ &=x\left[1+\frac12\cdot \frac 1{x^2}+\frac {\frac 12\cdot -\frac 12}{1\cdot 2}\frac1{4x^4}+\cdots \right]\\ &=x+\frac 1{2x}-\frac 1{8x^3}+\cdots \end{align}$$

Hypergeometricx
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