Inequality 12: Prove that: $\sqrt{1+\dfrac{bc }{a}}+\sqrt{1+\dfrac{ca }{b}}+\sqrt{1+\dfrac{ab }{c}}\ge 2\sqrt{3}.$

Let $a, b ,c$ be positive real number such that  $a+b+c=1$ , Prove that:

$\sqrt{1+\dfrac{bc }{a}}+\sqrt{1+\dfrac{ca }{b}}+\sqrt{1+\dfrac{ab }{c}}\ge 2\sqrt{3}.$

Solution

We will show that

$\left(1+\dfrac{bc}{a}\right)\left(1+\dfrac{ca}{b}\right)\left(1+\dfrac{ab}{c}\right) \ge \dfrac{64}{27}$

Note that

$\prod_{cyc}\left(1+\dfrac{bc}{a}\right) =\prod_{cyc}\dfrac{a(a+b+c)+bc}{a}$  $=\prod_{cyc}\dfrac{(a+b)(a+c)}{a}$ 
$=\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc}$

Again using the well known inequality

$(b+c)(c+a)(a+b) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca)$

We get that

$\prod_{cyc}\left(1+\dfrac{bc}a\right)$
$=\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64(a+b+c)^2(ab+bc+ca)^2}{81abc}$
$\ge \dfrac{64.3abc(a+b+c)}{81abc}=\dfrac{64}{27}$