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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