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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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On an infinite summation

Let \{x_n\}_{n=1}^{\infty} be a sequence of real numbers. Compute:

    \[\mathcal{V} = \sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n\]

Solution

First and foremost we set a_n = \sin^2 x_n and it is obvious that 0 \leq a_n \leq 1. We are making use of probabilistic methods. Suppose than an infinite number of coins are flipped. Let a_n be the probability that the n -th coin toss lands heads and let us consider the first time heads comes up. Then a_n \prod \limits_{k=1}^{n-1} (1 -a_k) is the probability that the first head appears in the n – th flip and \prod \limits_{n=1}^{\infty} (1-a_n) is the probability that all flips come up tails. Thus,

    \[\sum_{n=1}^{\infty} \sin^2 x_n \prod_{k=1}^{n-1} \cos^2 x_k + \prod_{n=1}^{\infty} \cos^2 x_n=1\]

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Trigonometric equality

Prove that in any triangle ABC it holds that

    \[\sum \sqrt{\frac{\sin A}{\sin B \sin C}} = \sqrt{\frac{2R}{r} \sum \sin A}\]

where R denotes the circumradius and r the inradius.

Solution

Using the law of sines we have that

    \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R\]

and if we denote \mathcal{A} the area of the triangle then

    \[\frac{r \left ( a + b + c \right )}{2} = \mathcal{A} = \frac{abc}{4R}\]

Thus,

    \begin{align*} \sum \sqrt{\frac{\sin A}{\sin B \sin C}} &= \sum \sqrt{\frac{a}{2R} \cdot \frac{2R}{b} \cdot \frac{2R}{c}} \\ &=\sum \sqrt{\frac{ a}{bc} \cdot 2 R} \\ &=\sum \sqrt{\frac{a}{bc} \cdot \frac{abc}{2\mathcal{A}}} \\ &=\frac{a+b+c}{\sqrt{2 \mathcal{A}}} \\ &= \sqrt{\frac{a+b+c}{\frac{2\mathcal{A}}{a+b+c}}} \\ &= \sqrt{\frac{a+b+c}{r}} \\ &= \sqrt{\frac{1}{r} \sum a} \\ &= \sqrt{\frac{2R}{r} \sum \frac{a}{2R}} \\ &= \sqrt{\frac{2R}{r} \sum \sin A} \end{align*}

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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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