Schur's inequality is a classical inequality that relates three non-negative real numbers.
For non-negative real numbers and and for a positive real number ,
with equality if and only if , or if two of , , are equal and the third is 0.
When the following special case arises:
Note that the expression in the left-hand side of Schur's inequality above is symmetric in the variables and Then, without loss of generality, suppose Factoring gives
Note that and Thus, the quantity inside the square brackets is non-negative. is also non-negative, so the left product is non-negative. Meanwhile, is non-negative, and and are both non-positive. Thus, the right product is non-negative. Hence, the entire expression is greater than or equal to zero.
, , and are non-negative reals such that . Prove that
Multiplying out and homogenizing, we want to prove
Recalling that when , Schur's inequality gives us , the inequality follows. In particular, due to the excess on the LHS, equality can only be achieved in the second condition, which implies all equality solutions are permutations of .
Let , and be positive reals such that , and . Prove that
We do the substitution , , , noting that (This is the incircle substitution including the degenerate case.) Now, the expression becomes
which is true by Schur's. Equality occurs when , or if two of , and are equal and the other is 0. Thus, all equality cases are of the form or are permutations of .