<s>[INST] Prove that $x+\frac{1}{2x}-\frac{1}{8x^3}<\sqrt{x^2+1}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!
 [/INST] 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
$$
\begin{align}
x+\frac1{2x}-\frac1{8x^3}
&=\frac{8x^4+4x^2-1}{8x^3}\\
&\le\frac{x^2+4x^2-1}{8x^3}\\
&\le\frac{\frac58-1}{8x^3}\\
&=-\frac3{64x^3}\\[4pt]
&\lt-\frac3{\sqrt8}
\end{align}
$$
</s>