# Bergstrom's Inequality and Applications

### Bergström's Inequality

The inequality below is often considered as a special form (Engel's form, after A. Engel) of Cauchy-Schwarz (Cauchy-Bunyakowski-Schwarz) inequality, but due to its numerous applications won a designation of its own. It was referred to in the classic book Inequalities by E. F. Beckenbach and R. Bellman, (Springer, 1961). In recent years it became better known as Titu's Inequality in honor of Titu Andreescu.

There are more accurate results.

### Proof of Bergström's Inequality

\displaystyle\begin{align} \left(\sum_{i=1}^{n}x_i\right)^2 &= \left(\sum_{i=1}^{n}\frac{x_i}{\sqrt{a_i}}\cdot\sqrt{a_i}\right)^2\\ &\le\sum_{i=1}^{n}\left(\frac{x_i}{\sqrt{a_i}}\right)^2\cdot\sum_{i=1}^{n}(\sqrt{a_i})^2\\ &=\sum_{i=1}^{n}\frac{x_i^2}{a_i}\sum_{i=1}^{n}a_i, \end{align}

which is equivalent to Bergström's inequality.

### Applications

1. Simple Nameless Inequality

Solution

2. Dorin Marghidanu's Example

3. Dorin Marghidanu's Second Example

If $a_1,a_2,\ldots,a_n\gt 1\,$ and $\sigma\in S_n\,$ (symmeric group), then

$\displaystyle \sum_{k=1}^n\frac{a_k^2}{a_{\sigma (k)}-1}\ge 4n.$

Solution

4. Nesbitt's Inequality

For $a,b,c\gt 0,$

$\displaystyle\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}$

Nesbitt's inequality has been shown by rearrangemnt elsewhere. Here we appeal to the rearrangement but after applying Bergström's inequality:

\displaystyle\begin{align} \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&=\frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}\\ &\ge\frac{(a+b+c)^2}{2(ab+bc+ca)}. \end{align}

Then suffice it to show that $\displaystyle\frac{(a+b+c)^2}{2(ab+bc+ca)}\ge\frac{3}{2}.\;$ This is equivalent to $a^2+b^2+c^2\ge ab+bc+ca\;$ which is true by rearrangement.

5. 36 IMO, Problem 2

Let $a,b,c,$ be positive real numbers such that $abc=1.\;$ Prove that

$\displaystyle\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge 1.$

With the substitution $\displaystyle a=\frac{1}{x},\;$ $\displaystyle b=\frac{1}{y},\;$ $\displaystyle c=\frac{1}{z},\;$ the required inequality is transformed into

$\displaystyle \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{3}{2}.$

With Bergström's inequality we get

\displaystyle\begin{align} \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}&\ge\frac{3}{2}\\ &\ge\frac{(x+y+z)^2}{2(x+y+z)}\\ &=\frac{x+y+z}{2}\\ &\ge \frac{3(xyz)^{1/3}}{2}\\ &=\frac{3}{2} \end{align}

because $xyz=1.$

6. Dimitris Kastriotis' Question

Kunihiko Chikaya shared his solution to a question posted by Dimitris Kastriotis on facebook.

Assume $a,b,c,$ are positive real numbers such that $a^2+b^2+c^2=3.\;$ Prove that

$\displaystyle\frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ca}\ge 1.$

Solution employs the obvious inequality $a^2+b^2\ge 2ab\,$ and one application of Bergstrom's inequality:

\displaystyle\begin{align} \frac{1}{1+2ab}+\frac{1}{1+2bc}+\frac{1}{1+2ca} &\ge \frac{1}{1+a^2+b^2}+\frac{1}{1+b^2+c^2}+\frac{1}{1+c^2+a^2}\\ &\ge\frac{(1+1+1)^2}{3+2(a^2+b^2+c^2)}\\ &=\frac{9}{3+2\cdot 6}\\ &=1. \end{align}

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