Prive that for a,b,c >0 and real

a/(b+c) + b/(c+a) + c/(a+b) >= 3/2

Consider a/(b+c) + b/(c+a) + c/(a+b) + 3...... which is

(a+b+c)[ 1/a+b + 1/b+c + 1/c+a ]

Now apply AM≥HM on for the expression in square bracket, and you will get

1/a+b + 1/b+c + 1/c+a ≥ 9/[2(a+b+c)].

So (a+b+c)[ 1/a+b + 1/b+c + 1/c+a ] ≥ 9/2

Or the given expression ≥ 9/2 - 3 = 3/2.

Hence given exp ≥ 3/2.

And that solves the problem.

Hope that helps.

All the best.

Regards,

Ashwin (IIT Madras).