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For every real positive number $a,b,c$ such that $ab+bc+ca=1$, how to prove that:

$$\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2\ge 16$$

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3 Answers 3

up vote 4 down vote accepted

This is problem J57 of mathematical reflections, vol.2007, issue 4;

$a^2+b^2+c^2 \ge ab+bc+ca =1$

$2\left(\frac ab+\frac bc+\frac ca\right)\ge6$ (as BenjaLim does)

It remains $\frac 1{a^2} + \frac 1{b^2} + \frac 1{c^2} \ge 9$ which follows by (use AGM) $\frac 3{(abc)^{2/3}}\ge9$ or $(abc)^{2/3}\le \frac13.$ This is a consequence of $ab+bc+ca=1$ since always by AGM $1= ab+bc+ca\ge 3(abc)^{2/3}$

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By Cauchy-Schwarz inequality,

$$\begin{eqnarray*} & & \left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2 \\ &=& \sqrt{\left(a+\frac{1}{b}\right)^2+\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2}\sqrt{\left(b+\frac{1}{c}\right)^2+\left(c+\frac{1}{a}\right)^2+\left(a+\frac{1}{b}\right)^2}\\ &\geq& \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)+\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)+\left(c+\frac{1}{a}\right)\left(a+\frac{1}{b}\right)\\ &=&3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}. \end{eqnarray*}$$

Now using the condition $ab+bc+ca = 1$,

$$\begin{eqnarray*} & & 3+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+ab+bc+ca+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \\ & = & \left(\frac{1}{ab}+9ab\right)+\left(\frac{1}{bc}+9bc\right)+\left(\frac{1}{ca}+9ca\right)+\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) - 5 \\ & \geq & 6+6+6+3-5 = 16. \end{eqnarray*}$$

We can also show that this inequality is strict, we can just put $(a, b, c) = \left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$.

(This minimizing case justifies our choice of $9ab+9bc+9ca$, combined with the equality condition of AM-GM inequality, and also of CS inequality.)

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Expanding the left hand side, we find that because for all non-zero real numbers $x$, $x^2 + x^{-2} \geq 2$ that

$$\left(a + \frac{1}{a}\right)^2 + \ldots \left(c + \frac{1}{c}\right)^2 \geq \frac{2a}{b} + \frac{2b}{c} + \frac{2c}{a} + 6.$$

Can you complete the argument?

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