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Tag Archives: Inequality

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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Root inequality

Let a, b, c, d be positive real numbers satisfying the following equality

    \[\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} =4\]

Prove that

\displaystyle \sqrt[3]{\frac{a^3+b^3}{2}} + \sqrt[3]{\frac{b^3+c^3}{2}} + \sqrt[3]{\frac{c^3+d^3}{2}} + \sqrt[3]{\frac{d^3+a^3}{2}} \leq 2 \left ( a+b+c+d \right ) -4

Solution

We begin by stating a lemma:

Lemma: Let a, b be positive real numbers, then:

    \[\frac{a+b}{2} \leq \sqrt[3]{\frac{a^3+b^3}{2}} \leq \frac{a^2+b^2}{a+b}\]

Now, making use of the lemma we have that:

    \begin{align*} \sum \sqrt[3]{\frac{a^3+b^3}{2}} &\leq \sum \frac{a^2+b^2}{a+b} \\ &=\sum \left ( a+b \right )-\sum \frac{2ab}{a+b} \\ &= 2\left ( a+b+c+d \right )- 2\sum \frac{1}{\frac{1}{a}+\frac{1}{b}} \end{align*}

Making use of the Cauchy – Schwartz inequality we have that

    \begin{align*} \sum \frac{1}{\frac{1}{a}+\frac{1}{b}} & \geq \frac{\left ( 1+1+1+1 \right )^2}{2\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d} \right )} \\ &=\frac{8}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}+ \frac{1}{d}} \\ &= 2 \end{align*}

The inequality now follows.

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A logarithmic inequality

Let x_1, x_2, \dots, x_n be n \geq 2 positive numbers other than 1 such that x_1^2+x_2^2+\cdots +x_n^2=n^3. Prove that

    \[\frac{\log_{x_1}^4 x_2}{x_1+x_2}+ \frac{\log_{x_2}^4 x_3}{x_2+x_3}+ \cdots + \frac{\log_{x_n}^4 x_1}{x_n+x_1} \geq \frac{1}{2}\]

Solution

The Engels form of the Cauchy – Schwartz inequality gives us:

    \begin{align*} \sum \frac{\log_{x_1}^4 x_2}{x_1+x_2} & \geq \frac{\left (\sum \log_{x_1}^2 x_2 \right )^2}{\sum (x_1+x_2)} \\ &= \frac{\left ( \sum \log_{x_1}^2 x_2 \right )^2}{2\sum x_1} \\ &\!\!\!\!\!\!\overset{\text{AM-GM}}{\geq } \frac{\left [ n \left (\prod \log_{x_1} x_2 \right )^{2/n} \right ]^2}{2\sum x_1} \\ &\!\!\!\!\overset{\text{C-B-S}}{\geq } \frac{n^2}{2n^2} \\ &= \frac{1}{2} \end{align*}

and the inequality is proven.

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Root inequality

Let a, b, c be three positive real numbers such that \sqrt{a} + \sqrt{b} + \sqrt{c}=1. Prove that

    \[\frac{\sqrt{a}}{a^2+2bc} + \frac{\sqrt{b}}{b^2+2ca} + \frac{\sqrt{c}}{c^2+2ab} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\]

Solution

By AM – GM we have,

    \begin{align*} \sum \frac{\sqrt{a}}{a^2+2bc} &\leq \sum \frac{\sqrt{a}}{a^2 + 2\left ( \frac{b^2+c^2}{2} \right )} \\ &= \frac{1}{a^2+b^2+c^2} \sum \sqrt{a}\\ &= \frac{1}{a^2+b^2+c^2} \end{align*}

However,

    \begin{align*} 1 &= \left (\sum \sqrt{a} \right )^2 \\ &=\left ( \sum \frac{a}{\sqrt{a}} \right )^2 \\ &\leq \left ( \sum a^2 \right ) \cdot \left ( \sum \frac{1}{a} \right ) \end{align*}

Hence \displaystyle \frac{1}{a^2+b^2+c^2} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c} and the exercise is complete.

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