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$f(x)=dfrac{ln x}{x}$
$ecause f^{''}(x)=dfrac{2ln x-3}{x^3}<0 herefore f^{'}(x)$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:$dfrac{x_1ln(x_2)-x_2ln(x_1)}{x_1x_2(x_2-x_1)}>dfrac{1-lnsqrt{x_1x_2}}{x_1x_2}$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.