# From China with love !

A Chinese mathematical olympiad problem and its solution:

Some body had once commented: The Silicon Valley is full of IC’s — Indians and Chinese! .

(My humble opinion: Perhaps, because of the strong emphasis on math in the two countries.)

Solve the inequality: $\log_{2}{(x^{12}+3x^{10}+5x^{8}+3x^{6}+1)}<1+\log_{2}{(x^{4}+1)}$

Solution:

As $1+\log_{2}{(x^{4}+1)}=\log_{2}{(2x^{4}+2)}$, and $\log_{2}{y}$ is monotonically increasing over $(0,+\infty)$, the given inequality is equivalent to :

$x^{12}+3x^{10}+5x^{8}+3x^{6}+1<2x^{4}+2$, that is,

$(\frac{1}{x^{2}})^{3}+2(\frac{1}{x^{2}})>x^{6}+3x^{4}+3x^{2}+1+2x^{2}+2$, which in turn, equals $((x)^{2}+1)^{3}+2(x^{2}+1)$.

Define: $g(t)=t^{2}+2t$. Then, we have: $g(\frac{1}{x^{2}})>g(x^{2}+1)$. Clearly, $g(t)$ is a monotonically increasing function; then we have

$\frac{1}{x^{2}}>x^{2}+1$

That is to say, $x^{4}+x^{2}-1<0$.

We obtain $x^{2}<\frac{-1+\sqrt{5}}{2}$. So the solution set is $(-\sqrt{\frac{-1+\sqrt{5}}{2}}, \sqrt{\frac{-1+\sqrt{5}}{2}})$.

Do you think you can solve this problem by another method? Please let me know.

— Nalin Pithwa.

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