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Logarithmic mean inequality

Let x, y>0 such that x \neq y. Prove that

    \[\sqrt{xy} \leq \frac{x-y}{\ln x- \ln y} \leq \frac{x+y}{2}\]

Solution

We are invoking the Hermite – Hadamard Inequality for the convex function f(x)=e^x \; , \; x \in \mathbb{R}. Thus,

    \begin{align*} f\left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{1}{\beta-\alpha} \int_{\alpha}^{\beta} f(x) \; \mathrm{d}x \leq \frac{f(\alpha)+f(\beta)}{2} &\Rightarrow \\ \exp \left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{1}{\beta-\alpha} \int_{\alpha}^{\beta} e^x \; \mathrm{d}x \leq \frac{e^\alpha+e^\beta}{2} &\Rightarrow \\ \exp \left ( \frac{\alpha+\beta}{2} \right ) \leq \frac{e^{\beta}-e^{\alpha}}{\beta-\alpha} \leq \frac{e^{\alpha}+e^{\beta}}{2}&\overset{\alpha=\ln x \;, \; \beta=\ln y}{=\! =\! =\! =\! =\! =\! =\!\Rightarrow } \\ \exp \left ( \frac{\ln x + \ln y}{2} \right ) \leq \frac{e^{\ln y} - e^{\ln x}}{\ln y- \ln x} \leq \frac{e^{\ln x} + e^{\ln y}}{2} \\ \exp \left ( \frac{\ln xy}{2} \right )\leq \frac{y-x}{\ln y- \ln x} \leq \frac{x+y}{2}&\Rightarrow \\ \sqrt{xy} \leq \frac{x-y}{\ln x - \ln y} \leq \frac{x+y}{2} \end{align*}

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