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Tag Archives: Inequality

Double inequality

In a triangle ABC prove that

    \[\frac{4}{9} \sum \sin B \sin C \leq \prod \cos \frac{B-C}{2} \leq \frac{2}{3} \sum \cos A\]

Vector inequality

Let \mathbb{R}^n be endowed with the usual product and the usual norm. If v = (x_1, x_2, \dots, x_n) \in \mathbb{R}^n then we define \sum v = x_1 +x_2 + \cdots + x_n. Prove that

    \[\left \| v \right \|^2 \left \| w \right \|^2 \geq \left ( v \cdot w \right )^2 + \frac{\left ( \left \| v \right \| \left | \sum w \right | - \left \| w \right \| \left |\sum v \right | \right )^2}{n}\]

Solution ( Robert Tauraso )

We will show the more general inequality

    \[\left ( \left \| v \right \|^2 \left \| w \right \|^2 - \left ( v \cdot w \right )^2 \right ) \left \| u \right \|^2 \geq \left \| \left ( w, u \right ) v - \left ( v, u \right ) w \right \|^2\]

where u \in \mathbb{R}^{n}. Taking u=(1, 1 , \dots, 1) we get the requested inequality. If v, w are linearly dependent then \left \| v \right \|^2 \left \| w \right \|^2 = \left ( v \cdot w \right )^2 the inequality holds. We assume now that v and w are linearly independent. Then u = \alpha v + \beta w + z where \alpha, \beta \in \mathbb{R} and z \perp vz \perp w. Moreover,

    \[\left\{\begin{matrix} \left ( v, u \right ) & = & \alpha \left \| v \right \|^2 + \beta \left ( v, w \right ) \\\\ \left ( w, u \right ) & = & \alpha \left ( v, w \right ) + \beta \left \| w \right \|^2 \end{matrix}\right.\]

and by solving the linear system we find

    \[\alpha = \frac{\left ( v, u \right ) \left \| w \right \|^2 - \left ( w, u \right ) \left ( v, w \right )}{\left \| v \right \|^2 \left \| w \right \|^2 - \left ( v, w \right )^2} \quad , \quad \beta = \frac{\left ( w, u \right )\left \| u \right \|^2 - \left ( v, u \right )\left ( v, w \right )}{\left \| v \right \|^2 \left \| w \right \|^2 - \left ( v, w \right )^2}\]

Hence,

\begin{aligned} \left ( \left \| v \right \|^2 \left \| w \right \|^2 - \left ( v \cdot w \right )^2 \right ) \left \| u \right \|^2 &= \left ( \left \| v \right \|^2 \left \| w \right \|^2 - \left ( v \cdot w \right )^2 \right ) \left ( \left \| \alpha v + \beta w \right \|^2 + \left \| z \right \|^2 \right ) \\ &\geq \left ( \left \| v \right \|^2 \left \| w \right \|^2 - \left ( v \cdot w \right )^2 \right ) \left ( \left \| \alpha v + \beta w \right \|^2 \right ) \\ &= \left ( \left \| v \right \|^2 \left \| w \right \|^2 - \left ( v \cdot w \right )^2 \right ) \left ( \alpha^2 \left \| v \right \|^2 + \beta^2 \left \| w \right \|^2 + 2\alpha \beta \left ( v, w \right ) \right ) \\ &= \left ( w, u \right )^2 \left \| v \right \|^2 + \left ( v, u \right ) \left \| w \right \|^2 - 2 \left ( w, u \right ) \left ( v,u \right ) \left ( v, w \right ) \\ &= \left \| \left ( w, u \right ) v - \left ( v, u \right )w \right \|^2 \end{aligned}

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Double inequality

Prove the following double inequality, where the sum and product are cyclic over the angles A, B, C of a triangle

    \[\sum \sin^2 A \leq 2 + 16 \prod \sin^2 \frac{A}{2} \leq \frac{9}{4}\]

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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