Let a, b, c >= 0 be side lengths of a triangle satisfying the claim. We have that sqrt(a) + sqrt(b) = sqrt(c) <=> (sqrt(a) + sqrt(b))^2 = c <=> a + b + 2*sqrt(a*b) = c, but by the triangle inequality, a+b >= c. Hence, the original equality holds if and only if 2*sqrt(a*b) = 0 <=> at least one of a, b are zero. WLOG we may assume that a=0. Now apply the equality again, but with a, b, c cyclically permuted: sqrt(b) + sqrt(c) = sqrt(a) = 0, since a=0. Consequently a=b=c=0. (Proof is for ANY triangle!) AIDIR?