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

Question

prove that; a/(b+c) +b/(a+c) +c/(a+b) >= 3/2

Answer 1

Let a=b=c then the inequality is true, because each fraction evaluates to ½ and the sum is 3/2.

The inequality is symmetrical in a, b and c.

Consider any two of the variables being equal: Let a=b. Let c<a (so c<b) and let c=a-x=b-x where x>0.

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

2a/(2a-x)+(a-x)/(2a)=(2a-x+x)/(2a-x)+½-x/(2a)=

1+x/(2a-x)+½-x/(2a)=(3/2)+(2ax-2ax+x²)/[2a(2a-x)]=

(3/2)+½x²/[a(2a-x)]. This expression is greater than 3/2 only when a(2a-x)>0.

Therefore (1) a>0 and 2a-x>0 or (2) a<0 and 2a-x<0.

Consider (1), then 0<x<2a. Let a=b=1, then 0<x<2; but if x≥2, let x=3 (c=-2) and let's see what the sum of the fractions is:

1/(1-2)+1/(1-2)-2/2=-1-1-1=-3. So the inequality is false because -3<3/2.

[If x=2, c=-1 and a+c=b+c=0 and the fraction sum can't be evaluated.]

Therefore, as it stands, the proposed inequality is false for general a, b, c, since we have an example (one of many) when a=b=1 and c=-2.

If a, b, c all have to be positive then, since c=a-x, a-x>0 and x<a, which implies that x<2a because a<2a. In this case the inequality is always satisfied. The question doesn't explicitly say that all the variables have to be positive.