Key idea. A common denominator reveals a weighted square.

1. Subtract the right side and use the positive common denominator \(ab(a+b)\):

   \[\frac{x^2}{a}+\frac{y^2}{b}-\frac{(x+y)^2}{a+b}=\frac{b(a+b)x^2+a(a+b)y^2-ab(x+y)^2}{ab(a+b)}.\]

2. Expand the numerator. The terms \(abx^2\) and \(aby^2\) cancel, leaving

   \[b^2x^2+a^2y^2-2abxy=(bx-ay)^2.\]

3. We have proved the identity

   \[\frac{x^2}{a}+\frac{y^2}{b}-\frac{(x+y)^2}{a+b}=\frac{(bx-ay)^2}{ab(a+b)}\ge0.\]

4. The denominator is positive, so equality holds exactly when \(bx-ay=0\), equivalently \(x/a=y/b\). No positivity condition on x or y was needed.

Remember. This two-term inequality can be applied repeatedly to handle three or more fractions.