# $\displaystyle \sum_{cyc} \dfrac{ab + bc + ca}{a(a + 2b)}\tag*{}$ If $a$, $b,$ and $c$ are positive reals, what is the minimum value of the above expression?

Observe that adding 3 to it, it is equal to

$\sum \frac{a(a+2b) + ab + bc + ca}{a(a+2b)} = \sum \frac{b(c+2a) + a(a+b+c)}{a(a+2b)} = (\sum \frac{b(c+2a)}{a(a+2b)}) + (\sum \frac{a+b+c}{a+2b})$

The first terms are at least 3 by AMGM. Examining the other term, let $s = \sum a$. Indeed, by Cauchy Schwarz:

$(\sum \frac{s}{a+2b})(\sum s(a + 2b)) \geq (\sum s)^2 = 9s^2$

which implies

$(\sum \frac{s}{a+2b}) \geq \frac{9s^2}{s \sum a+2b} = 3$

as desired.

## 2 Replies to “$\displaystyle \sum_{cyc} \dfrac{ab + bc + ca}{a(a + 2b)}\tag*{}$ If $a$, $b,$ and $c$ are positive reals, what is the minimum value of the above expression?”

1. Alex Wice says:

In what follows we aim to use as much basic mathematics as possible to achieve the result.

\begin{align*} \sum_{\mbox{cyc}}\frac{ab+bc+ca}{a(a+2b)} &= \frac{ab+bc+ca}{a(a+2b)}+\frac{ca+ab+bc}{c(c+2a)}+\frac{bc+ca+ab}{b(b+2c)} \\ &= (ab+bc+ca) \left( \frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \right) \\ &= abc\left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right) \left( \frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \right) \tag{1} \end{align*}

Now consider the denominators in the second pair of brackets, let us find their Arithmetic Mean (AM). To do this we will need their sum

\begin{align*}a(a+2b)+c(c+2a)+b(b+2c) &= a^{2}+b^{2}+c^{2}+2ab+2ac+2bc \\ &= (a+b+c)^{2}\tag{2}\end{align*}

From (2) the arithmetic mean of the denominators is now

$AM = \frac{(a+b+c)^{2}}{3} \tag{3}$

Another mean of the denominators could be the Harmonic Mean (HM) and as we shall see this is particularly helpful here

$HM = \frac{3}{\left( \frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \right) } \tag{4}$

Now a fundamental relationship exists between this pair of means, namely

$AM \ge HM \tag{5}$

or

$\frac{(a+b+c)^{2}}{3} \ge \frac{3}{\left( \frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \right) }$

which gives

$\frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \ge \frac{9}{(a+b+c)^{2}}\tag{6}$

Now using the same idea with the first pair of brackets in (1) we find

$\frac{a+b+c}{3} \ge \frac{3}{\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right )}$

or

$\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right ) \ge \frac{9}{a+b+c} \tag{7}$

If we use (6) and (7) in (1) we have

$\sum_{\mbox{cyc}}\frac{ab+bc+ca}{a(a+2b)} \ge \frac{81abc}{(a+b+c)^{3}} \tag{8}$

Now we have made extensive use of the Arithmetic Mean and the Harmonic mean, we turn now to the Geometric Mean (GM). A fundamental relationship being AM $\ge$ GM. So that

$\frac{a+b+c}{3} \ge \sqrt[3]{abc} \implies (a+b+c)^{3} \ge 27abc \tag{9}$

Using (9) in (8) we have

$\sum_{\mbox{cyc}}\frac{ab+bc+ca}{a(a+2b)} \ge \frac{81abc}{27abc} \ge 3 \tag{10}$

From (10) we see that the minimum possible value for the sum in (1) is $3$.

2. Willy Roentgen says:

For a = b = c =1 the expression is 1 (one)