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Tag Archives: Inequality

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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Nested radical inequality

Let n \in \mathbb{N}. Prove that

    \[\sqrt{2\sqrt[3]{3\sqrt[4]{4\cdots \sqrt[n]{n}}}}<2\]

Solution

The LHS is equal to 2^{1/2}3^{1/6} \cdots n^{1/n!} which by AM – GM is less or equal to

    \[\left( \frac{\sum_{k=2}^n (k/k!)}{\sum_{k=2}^n (1/k!)}\right)^{\sum_{k=2}^n (1/k!)} = \left(1 + \frac{1}{a_n} \right)^{a_n}\]

where a_n=\sum \limits_{k=2}^{n} \frac{1}{k!}. Since a_n \nearrow e-2 <2 it follows from Bernoulli inequality that \displaystyle \left(1 + \frac{1}{a_n} \right)^{a_n} <2.

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Root inequality

Let a, b, c be positive real numbers such that a+b+c=3. Prove that

    \[\sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} \leq \frac{3}{2} \sqrt[4]{\frac{1}{abc}}\]

Solution

Well if we apply AM-GM to (a^2, 1, 1, 1) we obtain

(1)   \begin{equation*} a^2+3 \geq 4 \sqrt{a} \end{equation*}

and similarly if we apply AM – GM to (b, b, b, c) we obtain

(2)   \begin{equation*} \frac{3b+c}{4} \geq \sqrt[4]{b^3c}\end{equation*}

We have successively,

\begin{aligned} \sqrt{\frac{b}{a^2+3}} + \sqrt{\frac{c}{b^2+3}} + \sqrt{\frac{a}{c^2+3}} &\overset{(1)}{\leq } \sqrt{\frac{b}{4\sqrt{a}}} + \sqrt{\frac{c}{4\sqrt{b}}} + \sqrt{\frac{a}{4\sqrt{c}}} \\ &=\frac{1}{2\sqrt[4]{abc}}\left ( \sqrt[4]{b^3c} +\sqrt[4]{c^3a}+ \sqrt[4]{a^3b}\right ) \\ &\overset{(2)}{\leq } \frac{1}{2 \sqrt[4]{abc}} \left ( \frac{3b+c}{4} + \frac{3c+a}{4} + \frac{3a+b}{4} \right )\\ &= \frac{3}{2} \sqrt[4]{\frac{1}{abc}} \end{aligned}

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Inequality with roots

Let a, b, c be positive real numbers. Prove that

    \[\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2\]

Solution

We apply the AM – GM inequality, thus:

    \begin{align*} \frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}} &=\frac{a}{b}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}} \\ &\geq 6\sqrt[6]{\frac{a}{b}\frac{b}{4c}\frac{c}{27a}} \\ &=\frac{6}{\sqrt[6]{4\cdot 27}} \end{align*}

Hence it suffices to prove that 3>\sqrt[6]{4\cdot 27} which holds because it is equivalent to 3^6> 4\cdot 27.

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