双变量的常规处理，但要注意齐次

\(f(x)=\dfrac{a}{x^2}+2\ln x\)

(1)求\(f(x)\)单调性

(2)若\(f(x)\)存在两个不同零点，证明：\(x_1f^{\prime}(x_1)+x_2f^{\prime}(x_2)>4\ln\dfrac{a}{2}+4\)

解
(1)当\(a\leq0\)，\(f(x)\)单调递增

当\(a>0\)
\(f^{\prime}(x)=\dfrac{2}{x}-\dfrac{2a}{x^3}=\dfrac{2(x^2-a)}{x^3}\)

则当\(x\in(0,\sqrt{a})\)时，\(f(x)\)减，\(x\in(\sqrt{a},+\infty)\)时，\(f(x)\)增

(2)由(1)，要使得\(f(x)\)有两个不同零点，则要有\(a>0,f(\sqrt{a})<0\)

则\(f(\sqrt{a})=1+\ln a<0\)，即\(0<a<e^{-1}\)

不妨设\(x_1<x_2\)

计算\(xf^{\prime}(x)=\dfrac{2(x^2-a)}{x^2}\)

则\(x_1f^{\prime}(x_1)+x_2f^{\prime}(x_2)=\dfrac{2(x_1^2-a)}{x_1^2}+\dfrac{2(x_2^2-a)}{x_2^2}=4-\dfrac{2a}{x_1^2}-\dfrac{2a}{x_2^2}\)

而\(\dfrac{a}{x_1^2}+2\ln x_1=0,\dfrac{a}{x_2^2}+2\ln x_2=0\)

从而\(x_1f^{\prime}(x_1)+x_2f^{\prime}(x_2)=4+4\ln x_1x_2\)

则原不等式为\(\ln x_1x_2>\ln\dfrac{a}{2}\)

即证\(x_1x_2>\dfrac{a}{2}\)

因\(\begin{cases}\dfrac{a}{x_1^2}+2\ln x_1=0\\ \\ \dfrac{a}{x_2^2}+2\ln x_2=0\end{cases}\)

两式相减有\(\dfrac{a}{x_1^2}-\dfrac{a}{x_2^2}=2(\ln x_1-\ln x_2)\)

整理有\(\dfrac{x_2^2-x_1^2}{\ln x_2-\ln x_1}=\dfrac{2x^2_1x^2_2}{a}\)

则要证\(x_1x_2>\dfrac{a}{2}\)

则证：\(\dfrac{x_2^2-x_1^2}{\ln x_2-\ln x_1}=\dfrac{2x^2_1x^2_2}{a}>x_1x_2\)

即证：\(\dfrac{x_2^2-x_1^2}{x_1x_2}>\ln \dfrac{x_2}{x_1}\)

即证：\(\dfrac{x_2}{x_1}-\dfrac{x_1}{x_2}>\ln\dfrac{x_2}{x_1}\)

令\(\dfrac{x_2}{x_1}=t\)，即证：

\(t-\dfrac{1}{t}>\ln t,t>1\)

这是个经典放缩，略.

得证！