prove that a^2/(2a^2+bc) + b^2/(2b^2+ca) + c^2/(2c^2 + ab) =1

Question

a,b,c are non zero real numbers such that a+b+c=0

Answer 1

Since a+b+c=0, c=-a-b. Substituting for c in the first two terms:

a²/(2a²-ab-b²)+b²/(2b²-ab-a²).

The symmetry in the denominators suggest that a-b is a factor (because putting a=b reduces the denominator to zero) in each case, so we can divide by this factor to find the other factor:

a²/((a-b)(2a+b))-b²/((a-b)(a+2b)).

So we have:

(a²(a+2b)-b²(2a+b))/((a-b)(2a+b)(a+2b))=

(a³+2a²b-2ab²-b³)/((a-b)(2a+b)(a+2b)).

This time the numerator factorises as a-b is a factor because of symmetry:

(a-b)(a²+3ab+b²), so (a-b) is a common factor in this fraction:

(a²+3ab+b²)/((2a+b)(a+2b)).

The numerator is the same as (a+b)²+ab=c²+ab.

Also, (2a+b)(a+2b)=(a-c)(b-c) when we substitute a+b=-c.

Now we can add this result to the third term:

(c²+ab)/((a-c)(b-c))+c²/(2c²+ab).

Expand (a-c)(b-c)=ab-c(a+b)+c²=ab+c²+c²=ab+2c².

So we have:

(c²+ab)/(ab+2c²)+c²/(2c²+ab)=(2c²+ab)/(2c²+ab)=1 QED.

Note that if one of a, b or c is zero in the original expression the result is also 1 (½+½=1).