解法六
赵斌，杭二中数学教师，IMO金牌教练，毕业于北京大学数学科学院，阿里巴巴全球数学竞赛奖获得者。
首先令
[imath]x=1[/imath]
得,
[imath]\sqrt{2} \leq \frac{1}{2 a},[/imath]
故得
[imath]a \in\left(0, \frac{\sqrt{2}}{4}\right][/imath]
.
下证当
[imath]a \in\left(0, \frac{\sqrt{2}}{4}\right), x \in\left[\frac{1}{e^{2}},+\infty\right) .[/imath]
必有，
[math]
f(x)=a \ln x+\sqrt{1+x} \leq \frac{\sqrt{x}}{2 a}
[/math]
首先由基本不等式
[imath]\ln t \le t-1 .t>0[/imath]
，得
[imath]\ln x =4 \ln x ^\frac14 \le 4 (x ^\frac14-1)[/imath]
，下证
[math]
4 a\left(x^{\frac{1}{4}}-1\right) \sqrt{1+x} \leq \frac{\sqrt{x}}{2 a}
[/math]
记
[imath]x^{\frac{1}{4}}=u \geq \frac{1}{\sqrt{\mathrm{e}}},[/imath]
故上式等价于
[math]
\begin{array}{c}
\sqrt{2} u^{2}-\sqrt{1+u^{4}}-\sqrt{2}(u-1) \geq(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right) \\
\Longleftrightarrow \frac{u^{4}-1}{\sqrt{2} u^{2}+\sqrt{1+u^{4}}}-\sqrt{2}(u-1) \geq(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right) \\
\Longleftrightarrow(u-1)\left(\frac{u^{3}+u^{2}+u+1}{\sqrt{2} u^{2}+\sqrt{1+u^{4}}}-\sqrt{2}\right) \geq(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right) \\
\Longleftrightarrow (u-1) \left(\frac{u^{3}+u^{2}+u+1}{\sqrt{2} u^{2}+\sqrt{1+u^{4}}}-\sqrt{2}\right) \geq (\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right) \\
\Longleftrightarrow \frac{(1-u)^{4}\left(u^{3}+u^{2}+u+1\right)}{\left(\sqrt{2} u^{2}+\sqrt{1+u^{4}}\right)\left(u^{3}-u^{2}+u+1+\sqrt{2\left(1+u^{4}\right)}\right)} \geq(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right)
\end{array}
[/math]
当
[imath]u \geq \frac{-1+\sqrt{5}}{2}[/imath]
时,有
[imath]1-u-\frac{u^{2}}{2 \sqrt{2} a} \leq 1-u-u^{2} \leq 0,[/imath]
故上式显然成立.
下设
[imath]u \in\left[\frac{1}{\sqrt{e}}, \frac{-1+\sqrt{5}}{2}\right],[/imath]
此时我们有
[math]
\frac{u^{3}+u^{2}+u+1}{\left(\sqrt{2} u^{2}+\sqrt{1+u^{4}}\right)\left(u^{3}-u^{2}+u+1+\sqrt{2\left(1+u^{4}\right)}\right)} \geq \frac{1}{3}
[/math]
(因为
[imath]\left.u^{3}+u^{2}+u+1 \geq \sqrt{2} u^{2}+\sqrt{1+u^{4}}, u^{3}-u^{2}+u+1+\sqrt{2\left(1+u^{4}\right)} \leq u^{3}+\frac{5}{4}+\sqrt{2\left(1+u^{4}\right)}<3 .\right)[/imath]
以下证明,
[math]
\frac{1}{3} \cdot(1-u)^{4} \geq(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right)
[/math]
当
[imath]a \le 0.3[/imath]
时，我们有,
[imath]1-u-\frac{u^{2}}{2 \sqrt{2} a} \leq 1-\frac{1}{\sqrt{e}}-\frac{1}{2 \sqrt{2} \times 0.3 \times e}<0,[/imath]
从而上式显然成立.
下设
[imath]a \geq 0.3,[/imath]
此时我们有，
[math]
\frac{1}{3} \cdot(1-u)^{4} \geq \frac{1}{3}\left(1-\frac{-1+\sqrt{5}}{2}\right)^{4}
[/math]
而
[math]
(\sqrt{2}-4 a)\left(1-u-\frac{u^{2}}{2 \sqrt{2} a}\right) \leq(\sqrt{2}-1)\left(1-u-u^{2}\right) \leq(\sqrt{2}-1)\left(1-\frac{1}{\sqrt{e}}-\frac{1}{e}\right)
[/math]
而经计算可得,
[math]
\frac{1}{3}\left(1-\frac{-1+\sqrt{5}}{2}\right)^{4} \geq(\sqrt{2}-1.2)\left(1-\frac{1}{\sqrt{e}}-\frac{1}{e}\right)
[/math]
故完成了证明.综上所求得
[imath]a[/imath]
的取值范围为
[imath]\left(0, \frac{\sqrt{2}}{4}\right].[/imath]
注：本解答事实上只是进行了进了一步比较难想的放缩，即
[imath]\ln x =4 \ln x ^\frac14 \le 4 (x^\frac14-1)[/imath]
，这个放缩效果会比
[imath]\ln x \le x-1[/imath]
更佳.
