2015年全国高中数学联赛天津市预赛

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因为当 $x \in (0,1)$ 时，有 $f'(x) = 1 - \dfrac{1}{x^{2}} <0,$ 所以 $f(x)$ 在 $(0,1]$ 上严格单调递减．于是对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有$$2-\sqrt{2} = f(1) \leqslant f(x)< f\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\sqrt{2}}{2}.$$再由 $f(x)$ 在 $(0,1]$ 上严格单调递减知对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有$$f(f(x)) >f\left(\dfrac{\sqrt{2}}{2}\right) = \dfrac{\sqrt{2}}{2}.$$对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，$f(f(x)) <x$ 当且仅当$$\dfrac{1}{x} + \dfrac{1}{x+\dfrac{1}{x} - \sqrt{2}} - 2\sqrt{2} <0.$$这又等价于对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有$$2\sqrt{2}x^{3} -6x^{2} + 3\sqrt{2}x -1>0.$$令 $g(x)=2\sqrt{2}x^{3} -6x^{2} + 3\sqrt{2}x -1$，则 $g\left(\dfrac{\sqrt{2}}{2}\right) = 0$．因为对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有\[\begin{split} g'(x) &=6\sqrt{2}x^{2} - 12x +3\sqrt{2} \\&= 3\sqrt{2}(2x^{2} -2\sqrt{2}x +1) \\&=3\sqrt{2}(\sqrt{2}x -1)^2 >0, \end{split}\]所以 $g(x)$ 在 $ \left[\dfrac{\sqrt{2}}{2},1\right]$ 上严格单调递增．因此对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有 $g(x)>g\left(\dfrac{\sqrt{2}}{2}\right) = 0$．这样就证明了对任意 $x \in \left(\dfrac{\sqrt{2}}{2},1\right]$，有 $f(f(x)) <x$