已知$0<x_1<c<x_2<e^{\frac{3}{2}},$且$\dfrac{1-ln(c)}{c^2} = \dfrac{x_1ln(x_2)-x_2ln(x_1)}{x_1x_2(x_2-x_1)}$,
证明:$c^2<x_1x_2$
由题意,结合拉格朗日中值定理知:$f^{'}(c)=\dfrac{x_1ln(x_2)-x_2ln(x_1)}{x_1x_2(x_2-x_1)}$,其中$f(x)=\dfrac{\ln x}{x}$
$\because f^{''}(x)=\dfrac{2\ln x-3}{x^3}<0\therefore f^{'}(x)$单调递减.要证明$c^2<x_1x_2$只需证明:$f^{'}(c)>f^{'}(\sqrt{x_1x_2})$
即证明:$\dfrac{x_1ln(x_2)-x_2ln(x_1)}{x_1x_2(x_2-x_1)}>\dfrac{1-\ln\sqrt{x_1x_2}}{x_1x_2}$化简得
$(x_1+x_2)\ln(x_2)-(x_1+x_2)\ln(x_1)>2(x_2-x_1)$,令$t=\dfrac{x_2}{x_1}>1$,即证:$\ln t>\dfrac{2(t-1)}{t+1}$易知成立.