Q13/066) Prove that (a+b+c)/(a+b) + (a+b+c)/(b+c) + (a+b+c)/(c+a) > 9/2

a/(b+c) + b/(c+a)
= (ca + a^2 + b^2 + bc)/(b+c)(c+a)
>= (ca + 2ab + bc)/(b+c)(c+a) [since (a-b)^2 >= 0 implies a^2 + b^2 >= 2ab]
= {a(b+c) + b(c+a)}/(b+c)(c+a)

Thus, a/(b+c) + b/(c+a) >= a/(c+a) + b/(b+c)

Similarly, it can be proved that

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

Adding corresponding sides of the above three inequalities, we get

2{a/(b+c) + b/(c+a) + c/(a+b)} >= 3
i.e. a/(b+c) + b/(c+a) + c/(a+b) >= 3/2

adding one to each term on LHS and 3 on RHS

( 1 + a/(b+c)) + ( 1+ b/(c+a)) + ( 1+ c/(a+b)) >= 3/2 + 3

or (a+b+c)/(a+b) + (a+b+c)/(b+c) + (a+b+c)/(c+a) > 9/2

QED