# work 18 – ML user Colourfuldreams

Prove that for $a,b,c$ with sum 1 , we have the following inequality –

$\sum \sqrt{1 + \frac{bc}{a}} \ge 3\sqrt{2}$

Proof

The LHS is equivalent to –

$\sum \sqrt{\frac{(a+b)(a+c)}{a}}$

Let $a = \frac{1}{a} \cdots$

Therefore the condition is now –

$ab+bc+ca = abc$

The LHS is also transformed to –

$\sum \sqrt{\frac{(a+b)(a+c)}{ab+bc+ca}}$

Thus we have to prove that –

$\sum \sqrt{(a+b)(a+c)} \ge 3\sqrt{2(ab+bc+ca)}$

Squaring and using the fact that –

$\sqrt{(a+b)(a+c)} \ge a + \sqrt{bc}$ (By C-S)

we get the desired inequality.

Equality occurs for $a=b=c=\frac{1}{3}$