Mathematics - Quadratic Equation Question with Solution | TestHub

Using the $\mathrm{AM}\ge \mathrm{HM}$ of $\frac{1}{\mathrm{a}},\frac{1}{\mathrm{b}},\frac{1}{\mathrm{c}}$ we get, $\frac{\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}}{3}\ge \frac{3}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=\frac{3}{6}=\frac{1}{2}$. So, $\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}\ge \frac{3}{2}$.

Now,

$$\begin{array}{rl}& \frac{{\left(a+\frac{1}{b}\right)}^2+{\left(b+\frac{1}{c}\right)}^2+{\left(c+\frac{1}{a}\right)}^2}{3}\ge {\left(\frac{a+\frac{1}{b}+b+\frac{1}{c}+c+\frac{1}{a}}{3}\right)}^2\\ & \ge {\left(\frac{6+\frac{3}{2}}{3}\right)}^2={\left(\frac{15/2}{3}\right)}^2={\left(\frac{5}{2}\right)}^2=\frac{25}{4}\\ & \therefore {\left(a+\frac{1}{b}\right)}^2+{\left(b+\frac{1}{c}\right)}^2+{\left(c+\frac{1}{a}\right)}^2\ge \frac{75}{4}\end{array}$$