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}

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