证明下列不等式：
(1) 若$\dfrac{1}{\rm e}<x<y<1$，则$\dfrac yx<\dfrac{1+\ln y}{1+\ln x}$；
(2) $\left(\dfrac 2{1^4}+1\right)\left(\dfrac{2}{2^4}+1\right)\cdots \left(\dfrac{2}{n^4}+1\right)<{\rm e}^4$．
证明 (1) 由于$\dfrac yx<\dfrac{1+\ln y}{1+\ln x}$等价于$\dfrac{1+\ln x}x<\dfrac{1+\ln y}y$，于是只需要证明函数$f(x)=\dfrac{1+\ln x}x$在区间$\left(\dfrac{1}{\rm e},1\right)$上单调递增．函数$f(x)$的导函数$$f'(x)=\dfrac{-\ln x}{x^2},$$于是函数$f(x)$在$(0,1)$上单调递增，因此原命题得证．
(2) 等价于证明$$\ln\left(\dfrac 2{1^4}+1\right)+\ln\left(\dfrac{2}{2^4}+1\right)+\cdots +\ln\left(\dfrac{2}{n^4}+1\right)<4,$$而$\ln (x+1)<x$($x>0$)，于是有\[\begin{split} LHS&<\dfrac{2}{1^4}+\dfrac{2}{2^4}+\dfrac{2}{3^4}+\dfrac{2}{4^4}+\cdots +\dfrac{2}{n^4}\\&\leqslant 2+\dfrac 18+\dfrac{2}{1\cdot 2\cdot 3\cdot 4}+\dfrac 2{2\cdot 3\cdot 4\cdot 5}+\cdots +\dfrac{2}{(n-2)(n-1)n(n+1)}\\&=\dfrac{17}8+\dfrac 23\cdot \left[\dfrac{1}{1\cdot 2\cdot 3}-\dfrac{1}{(n-1)n(n+1)}\right]\\&<\dfrac{17}8+\dfrac 19<4,\end{split} \]从而原命题得证．
注 因为不等式相当宽松，也可以直接先将分母放缩成二次，得到结果，即\[\begin{split} LHS&<\dfrac{2}{1^4}+\dfrac{2}{2^4}+\dfrac{2}{3^4}+\dfrac{2}{4^4}+\cdots +\dfrac{2}{n^4}\\&\leqslant \dfrac{2}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+\cdots +\dfrac{2}{n^2}\\&<2\left(1+1-\dfrac 12+\dfrac 12-\dfrac 13+\cdots+\dfrac 1{n-1}-\dfrac 1n\right)\\&=4-\dfrac 2n<4,\end{split} \]