# Prove that "sqrt[a/(b+c)]+sqrt[b/(c+a)]+sqrt[c/(a+b)]>2" (if a,b,c >0) (?)

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### 1 Answer

`sqrt(a/(b+c))=sqrt(1+(a-b-c)/(b+c))`

`(1+(a-b-c)/(b+c))^(1/2)=sqrt(a/(b+c))`

`` Expand by binomial theorem

`sqrt(a/(b+c))=1+(1/2)(a-b-c)/(b+c)+..... (i)`

`Similarly we can write`

`sqrt(b/(a+c))=1+(1/2)(b-a-c)/(a+c)+..... (ii)`

`sqrt(c/(b+a))=1+(1/2)(c-b-a)/(b+a)+..... (iii)`

`add (i) (ii) and (iii)`

`sqrt(a/(b+c))+sqrt(b/(a+c))+sqrt(c/(a+b))=3+(1/2)(a-b-c)/(b+c)+...`

`sqrt(a/(b+c))+sqrt(b/(a+c))+sqrt(c/(a+b))>3`

`sqrt(a/(b+c))+sqrt(b/(a+c))+sqrt(c/(a+b))>2`

`If A>3 then A>2`

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