已知函式\(f(x)=\dfrac{\ln x+1}{x}.g(x)=\dfrac{e^x}{x}\)
(1)若對任意\(m,n\in\left(0,+\infty\right)\)都有\(f(m)\leq t\leq g(n)\)，求實數\(t\)的取值範圍.
(2)若對任意\(x_1,x_2\in(0,+\infty)\)且\(x_1\neq x_2,e^{x_2-x_1}=\dfrac{x_2^{x_1}}{x_1^{x_2}}\)，證明\(x_1^3+x_2^3>2\)
解
(1)\(f(x)\leq \dfrac{x-1+1}{x}=1,g(n)\geq \dfrac{ex}{x}=e\)
則\(1\leq t\leq e\)
(2)\(e^{x_2-x_1}=\dfrac{x_2^{x_1}}{x_1^{x_2}}\)取對數有
\(x_2-x_1=x_1\ln x_2-x_2\ln x_1\)，整理得\(\dfrac{1+\ln x_1}{x_1}=\dfrac{1+\ln x_2}{x_1}\)
即\(f(x_1)=f(x_2)\)，即\(x_1<1<x_2\)
現在說明\(x_1+x_2>2\)(極值點偏移做多了，應該想到的)
即說明\(x_2>2-x_1\)
構造\(h(t)=f(t)-f(2-t),t<1\)
即\(h^{\prime}(t)=-\dfrac{\ln x}{x}-\dfrac{\ln(2-x)}{(2-x)^2}=-\dfrac{\ln\left[(-(x-1)^2)+1\right]}{x^2}>0\)
則\(h(t)\)單調遞增，則\(h(t)<h(1)=0\)
則\(x_2>2-x_1\)即\(x_1+x_2>2\)
則\(x_1^3+x_2^3>x_1^3+(2-x_1)^3=8-12x_1+6x-1^2=6(x_1-1)^2+2>2\)