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Tag Archives: Inequality

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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Complex inequality

Prove that forall z \in \mathbb{C} and n \in \mathbb{N} it holds that

    \[\left | \left ( 1+z \right )^n -1 \right | \leq \left ( 1+\left | z \right | \right )^n -1\]

Solution

Well,

    \begin{align*} \left | \left ( 1+z \right )^n -1 \right | &= \left | \sum_{k=0}^{n} \binom{n}{k} z^k -1 \right | \\ &= \left | \sum_{k=1}^{n} \binom{n}{k} z^k\right |\\ &\leq \sum_{k=1}^{n} \binom{n}{k} \left | z^k \right | \\ &=\sum_{k=0}^{n} \binom{n}{k} \left | z \right |^n -1 \\ &= \left ( 1+\left | z \right | \right )^n -1 \end{align*}

Done!

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Value of parameter

This is a very classic exercise and can be dealt with various ways. We know the result in advance. Why? Because it is the Taylor Polynomial of the exponential function. Let us see however how we gonna deal with it with High School Methods.

Find the positive real number \alpha such that

    \[e^x \geq 1+x+\frac{x^2}{2} + \alpha x^3 \quad  \quad \text{forall} \;\; x \in \mathbb{R}\]

Solution

Define the function

    \[f(x) = \frac{1+x+\frac{x^2}{2}+\alpha x^3}{e^x} \quad , \quad x \in \mathbb{R}\]

and note that f(x) \leq 1 forall x \in \mathbb{R}. Clearly , f is differentiable and its derivative is given by

    \[f'(x)= \frac{x^2\left ( 6\alpha-1-2\alpha x \right )}{e^x} \quad , \quad x \in \mathbb{R}\]

It follows that \displaystyle f'(x) =0 \Leftrightarrow \left\{\begin{matrix} x &= &0 \\\\ x & = & \dfrac{6\alpha-1}{2\alpha} \end{matrix}\right.. Suppose that \frac{6\alpha-1}{2\alpha}>0. Then the monotony of f as well as the sign of f' is seen at the following table.

Rendered by QuickLaTeX.com

It follows then that \displaystyle f\left ( \frac{6\alpha-1}{2\alpha} \right )> f(0)=1. This is an obscurity due to the fact that f(x) \leq 1. Similarly, if we suppose that \frac{6\alpha-1}{2\alpha}<0. Hence

    \[\frac{6\alpha-1}{2\alpha}=0 \Leftrightarrow \alpha=\frac{1}{6}\]

For \alpha=\frac{1}{6} we easily see that the given inequality holds.

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Integral and inequality

Let f:[0,1] \rightarrow \mathbb{R} be a continuous function such that

(1)   \begin{equation*} \int_0^1 f(x) \, \mathrm{d}x = \kappa =\int_0^1 x f(x) \, \mathrm{d}x \end{equation*}

Prove that

    \[\int_0^1 f^2(x) \, \mathrm{d}x \geq 4\kappa^2\]

Solution

We note that

    \begin{align*} \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x &= \int_{0}^{1} f^2(x) \, \mathrm{d}x -12 \kappa \int_{0}^{1} x f(x) \, \mathrm{d}x + 36 \kappa^2 \int_{0}^{1} x^2 \, \mathrm{d}x \\ &=\int_{0}^{1} f^2(x) \, \mathrm{d}x - 12 \kappa \int_{0}^{1} f(x) \, \mathrm{d}x + \frac{36\kappa^2}{3} \\ &=\int_{0}^{1} f^2 (x) \, \mathrm{d}x - 12 \kappa^2 + 12 \kappa^2\\ &= \int_{0}^{1} f^2(x) \, \mathrm{d}x \end{align*}

On the other hand by Cauchy – Schwartz we have that

    \begin{align*} \int_{0}^{1} f^2(x) \, \mathrm{d}x &= \int_{0}^{1} \left ( f(x) - 6 \kappa x \right )^2 \, \mathrm{d}x \\ &\geq \left [ \int_{0}^{1} \left ( f(x) - 6 \kappa x \right ) \, \mathrm{d}x \right ]^2 \\ &=\left ( \kappa - 3 \kappa \right )^2 \\ &= 4\kappa^2 \end{align*}

The result follows.

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Inequality of a concave function

Let f:[0, 1] \rightarrow [0, +\infty) be a concave function. Prove that

    \[\int_0^1 x^2 f(x) \, \mathrm{d}x \leq \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x\]

Solution

Since f is concave , it holds that

    \[f\left ( ax + \left ( 1-a \right )y \right ) \geq a f(x) + \left ( 1-a \right ) f(y) \geq a f(x) \quad \text{forall} \;\; a, x, y \in [0, 1]\]

By setting y=0 and x=a we get that x f(x) \leq f \left ( x^2 \right ). Thus,

    \begin{align*} \int_{0}^{1} x^2 f(x) \, \mathrm{d}x &= \frac{1}{2} \int_{0}^{1} x f(x) 2x \, \mathrm{d}x \\ &\leq \frac{1}{2}\int_{0}^{1} f \left ( x^2 \right ) 2x \, \mathrm{d}x \\ &=\frac{1}{2} \int_{0}^{1} f \left ( x^2 \right ) \, \mathrm{d} \left ( x^2 \right ) \\ &= \frac{1}{2} \int_{0}^{1} f(x) \, \mathrm{d}x \end{align*}

and the exercise is complete.

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