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Tag Archives: Inequality

Gamma inequality

Let a ,b, c be three positive real numbers such that abc=1. Prove that:

    \[\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{\Gamma(b)}{1+b+bc} + \frac{\Gamma(c)}{1+c+ca} \geq 1\]

where \Gamma is Euler’s Gamma function.

Solution

We can rewrite the inequality as

    \begin{align*} \sum \frac{\Gamma\left ( a \right )}{1+a+ab} &= \frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a}{a} \frac{\Gamma(b)}{1+b+bc} + \frac{ab}{ab} \frac{\Gamma(c)}{1+c+ca} \\ &=\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a\Gamma(b)}{a+ab+abc} + \frac{ab\Gamma(c)}{ab+abc+abca} \\ &=\frac{\Gamma\left ( a \right )}{1+a+ab} + \frac{a \Gamma(b)}{1+a+ab} + \frac{ab \Gamma(c)}{1+a+ab} \\ &=\frac{\Gamma(a) + a \Gamma(b) + ab \Gamma(c)}{1+a+ab} \\ &\geq \Gamma \left ( \frac{1+a + ab}{1+a+ab} \right ) \\ &= 1 \end{align*}

since \Gamma is convex.

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Integral inequality

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

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

Prove that \int_0^1 f^2(x) \, \mathrm{d}x \geq 4 \kappa^2.

Solution

We have successively:

    \begin{align*} \int_0^1 f^2(x) \, \mathrm{d}x &= \int_0^1 (3x-1)^2 \, \mathrm{d}x \int_0^1 f(x)^2 \, \mathrm{d}x \\ & \geq \left( \int_0^1 (3x-1)f(x) \, \mathrm{d}x \right)^2 \\ &= (3k-k)^2 \\ &= 4k^2 \end{align*}

Since f is continuous  equality holds only if f(x)=C(3x-1) for some constant C. From the data we get C=2k.

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“Upper” bound

Let f:\mathbb{R}^+ \rightarrow \mathbb{R} be a function satisfying

    \[\left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a}\]

for all positive real numbers a and b. Prove that

    \[\left | f(1) - f(x) \right | \leq \left |\ln x \right | \quad \text{for all} \;\; x>0\]

Solution

For starters, let us assume that x>1. Dividing the interval (1, x) into n subintervals each of length h so that h=\frac{x-1}{n}. Thus,

    \begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \end{align*}

The inequality \left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a} implies that

    \[\left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \leq \frac{h}{1+kh}\]

Hence,

    \[\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}\]

The limit \displaystyle \sum_{k=0}^{n-1} \frac{h}{1+kh} exists and equals to \ln x. Hence , the inequality is proved for x>1.

Now, assume that x<1. Dividing the interval (x, 1) into n subintervals each of length h so that h=\frac{1-x}{n}. Thus,

    \begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \end{align*}

The inequality \left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a} implies that

    \[\left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \leq \frac{h}{1-kh}\]

Hence,

    \[\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}\]

The limit \displaystyle \sum_{k=0}^{n-1} \frac{h}{1-kh} exists and equals to -\ln x. Hence , the inequality is also proved for x<1. This completes the proof!

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Upper bound of max product

Let z_1, z_2 , \dots, z_n \in \mathbb{C} be the roots of the polynomial

    \[f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0 \in \mathbb{C}[x]\]

Prove that:

    \[\prod_{k = 1}^{n} \max (1,|z_k|) \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Solution

Suppose the roots of polynomial f are z_1, z_2, \dots, z_n where

    \[|z_1| \geq |z_2| \geq \cdots \geq |z_m| > 1 \geq |z_{m+1}| \geq \cdots \geq |z_n|\]

Let g(z)=z^nf \left(\frac{1}{z}\right) = 1 + a_{n-1}z + \cdots + a_0z^n. Then, the \{1/z_k\}_{1\leq k \leq m} are the zeros of g in the disk |z| \le r = 1-\epsilon < 1 where \epsilon is chosen such that g(re^{i\theta}) \neq 0 for \theta \in [0, 2\pi].

Jensen’s inequality implies that

    \begin{align*} \log|r^m z_1 \cdots z_m| = \frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d} \theta & \Leftrightarrow \\ |r^m z_1\cdots z_m| = \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) &\Leftrightarrow \\ \exp \left(\frac{1}{2\pi}\int_0^{2\pi} \log |g(re^{i\theta})|\,\mathrm{d}\theta\right) \le \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta \end{align*}

Applying Cauchy – Schwartz yields,

    \begin{align*} \frac{1}{2\pi}\int_0^{2\pi} |g(re^{i\theta})|\,\mathrm{d}\theta &\leq \frac{1}{2\pi}\left(\int_0^{2\pi}\,\mathrm{d}\theta\int_0^{2\pi} |g(re^{i\theta})|^2\,\mathrm{d}\theta\right)^{1/2} \\ &= \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2} \end{align*}

Therefore,

    \[\left|r^m z_1\cdots z_m\right| \leq \sqrt{1+|a_{n-1}|^2 + \cdots + |a_{0}|^2}\]

Letting r\rightarrow 1 and \epsilon \rightarrow 0 we get the result.

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Offset logarithmic integral inequality

Prove that

    \[\int_2^{e+1} \frac{\mathrm{d}t}{\ln t} < e\]

Solution

We have successively:

    \begin{align*} \hspace{-1em}\ln x \leq x-1 &\Rightarrow \ln \frac{1}{x} \leq \frac{1}{x}-1 \\ &\Rightarrow -\ln x \leq \frac{1}{x}-1 \\ &\Rightarrow \ln x \geq 1-\frac{1}{x} \\ &\Rightarrow \frac{1}{\ln x} \leq \frac{1}{1-\frac{1}{x}} \\ &\Rightarrow \int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < \int_2^{e+1} \frac{\mathrm{d}t}{1-\frac{1}{t}} \\ &\Rightarrow \mathbf{\int_{2}^{e+1} \frac{\mathrm{d}t}{\ln t} < e} \end{align*}

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