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

Observe that adding 3 to it, it is equal to

[math]\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})[/math]

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

[math](\sum \frac{s}{a+2b})(\sum s(a + 2b)) \geq (\sum s)^2 = 9s^2[/math]

which implies

[math](\sum \frac{s}{a+2b}) \geq \frac{9s^2}{s \sum a+2b} = 3[/math]

as desired.

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

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

    [math]\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*}[/math]

    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

    [math]\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*}[/math]

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

    [math]AM = \frac{(a+b+c)^{2}}{3} \tag{3}[/math]

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

    [math]HM = \frac{3}{\left( \frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \right) } \tag{4}[/math]

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

    [math]AM \ge HM \tag{5}[/math]

    or

    [math]\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) }[/math]

    which gives

    [math]\frac{1}{a(a+2b)}+\frac{1}{c(c+2a)}+\frac{1}{b(b+2c)} \ge \frac{9}{(a+b+c)^{2}}\tag{6}[/math]

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

    [math]\frac{a+b+c}{3} \ge \frac{3}{\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right )}[/math]

    or

    [math]\left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right ) \ge \frac{9}{a+b+c} \tag{7}[/math]

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

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

    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 [math]\ge[/math] GM. So that

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

    Using (9) in (8) we have

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

    From (10) we see that the minimum possible value for the sum in (1) is [math]3[/math].

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