prove that
(a/b)+(b/c)+(c/a) is greater than or equal to [(a+b+c)^2]/(ab+bc+ca)

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Solution

$Dear Student, Lets take \frac{{(a + b + c)}^{2}}{ab + bc + ca} = \frac{a^{2} + b^{2} + c^{2} + 2 (ab + bc + ca)}{ab + bc + ca} = \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} + 2 Now take \frac{a}{b}, \frac{b}{c}, \frac{c}{a} and using AM \geq GM, \frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq {(\frac{a}{b} \times \frac{b}{c} \times \frac{c}{a})}^{\frac{1}{3}} \geq 1 \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 and \frac{a^{2} + b^{2} + c^{2}}{ab + bc + ca} + 2 \geq 2 So combining above 2 result \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{{(a + b + c)}^{2}}{ab + bc + ca} Regards$

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