Mathematics - Quadratic Equation Question with Solution | TestHub

Using the weighted $\mathrm{AM}-\mathrm{GM}$ in equality we get,

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

$\begin{array}{c}\frac{b\cdot a+c\cdot b+a\cdot c}{b+c+a}\ge {\left(a^b\cdot b^c\cdot c^a\right)}^{\frac{1}{a+b+c}}\end{array}$

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

Adding these inequalities together we get,

$$\begin{array}{rlr}& \frac{a^2+b^2+c^2+2(ab+bc+ca)}{a+b+c}\ge \left(a^a\cdot b^b\cdot c^c\right)+\left(a^c{b}^a{c}^b\right)+\left(a^b{b}^c{c}^a\right)[\because a+b+c=1]& \end{array}$$

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