# Inequality with Constraint from Dan Sitaru's Math Phenomenon

### Statement ### Solution 1

Note that

\displaystyle\begin{align} 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}&\ge 2\sum_{cycl}\frac{\frac{3}{4}(a+b)^2}{a+b}\\ &= 2\sum_{cycl}\frac{3}{4}(a+b)\\ &=2\cdot\frac{3}{2}(a+b+c)\\ &=3(a+b+c)\\ &\ge (a+b+c)+2(a+b+c)\\ &\ge b+2c+20, \end{align}

because $a\ge c.\,$ This proves the right inequality. The left inequality is equivalent to

$\displaystyle 4a+3b+2c\ge 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b},$

which, in turn is equivalent to

\displaystyle\begin{align} \left(2a+b-\frac{2(a^2+ab+b^2)}{a+b}\right)&+\left(2b+c-\frac{2(b^2+bc+c^2)}{b+c}\right)\\ &+\left(2a+c-\frac{2(a^2+ac+c^2)}{a+c}\right)\ge 0, \end{align}

or,

$\displaystyle \frac{b(a-b)}{a+b}+\frac{b(b-c)}{b+c}+\frac{c(a-c)}{c+a}\ge 0,$

which is true because $a\ge b\ge c.$

### Solution 2

The left inequality is equivalent to

\displaystyle \begin{align} 4a+3b+2c &\ge 2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}\\ &=2\sum_{cycl}\frac{a^2+2ab+b^2}{a+b}-2\sum_{cycl}\frac{ab}{a+b}\\ &=2\sum_{cycl}(a+b)-2\sum_{cycl}\frac{ab}{a+b}\\ &=4(a+b+c)-2\sum_{cycl}\frac{ab}{a+b}, \end{align}

which can be rewritten as

$\displaystyle 2\sum_{cycl}\frac{ab}{a+b}\ge b+2c.$

Now note that

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

Also $\displaystyle \frac{2ca}{c+a}\ge c\,$ is equivalent to $2ca\ge c^2+ca,\,$ or $a\ge c,\,$ which is true. Now, adding this to

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

completes the proof of the left inequality. For the right inequality, observe that, as we just showed,

$\displaystyle 2\sum_{cycl}\frac{ab}{a+b}\ge b+2c.$

Thus suffice it to prove that

$\displaystyle 2\sum_{cycl}\frac{a^2+b^2}{a+b}\ge 20.$

This is indeed so due to Bergstrom's inequality:

\displaystyle\begin{align}2\sum_{cycl}\frac{a^2+b^2}{a+b}&=2\sum_{cycl}\frac{a^2}{a+b}+2\sum_{cycl}\frac{b^2}{a+b}\\ &\ge 2\frac{(a+b+c)^2}{2(a+b+c)}+2\frac{(a+b+c)^2}{2(a+b+c)}\\ &=2(a+b+c)=20. \end{align}

### Solution 3

Let $\displaystyle f=2\sum_{cycl}\frac{a^2+ab+b^2}{a+b}.\,$ We reexpress:

$\displaystyle f=2\sum_{cycl}\left(\frac{(a+b)^2-ab}{a+b}\right)=2\left(20-\sum_{cycl}\frac{ab}{a+b}\right).$

Let us further establish that from the assumtions $a\ge b\ge c\gt 0\,$ and $a+b+c=10,\,$ it is necessary that $\displaystyle a\ge \frac{10}{3}\,$ and $\displaystyle c\le\frac{10}{3}.$

We have the right side inequality:

$\displaystyle b+2c+20\le 20+\left(1-\frac{10}{3}\right)+\frac{10}{3}\le 30.$

We also have $f\ge 30,\,$ for, $\displaystyle \frac{ab}{a+b}\le\frac{1}{4}(a+b)\,$ from which

$\displaystyle \sum_{cycl}\frac{ab}{a+b}\le\frac{1}{2}(a+b+c)=5.$

For the left inequality, we have

$\displaystyle \frac{ab}{a+b}+\frac{cb}{b+c}+\frac{ca}{c+a}\ge\frac{ab}{2a}+\frac{bc}{2b}+\frac{ca}{2a}=\frac{b}{2}+c,$

so $f\le 40-b-2c.\,$ The left side inequality becomes $b+2a+20\ge 40-b-2c,\,$ i.e., $2(a+b+c)\ge 20\,$ whcih is true.

### Acknowledgment

The above problem, originally from his book Math Phenomenon, has been posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Diego Alvariz; Solution 3 is by N. N. Taleb. 