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SCHUR不等式例题


a, b, c, are non-negative reals such that a+b+c=1. Prove that a3+b3+c3+6abc≥1/4 Solution. Multiplying by 4 and homogenizing, 4×(a3+b3+c3+6abc)≥4×(1/4) 4×(a3+b3+c3+6abc)≥1 we seek 4a3+4b3+4c3+

24abc≥(a+b+c)3 (a+b+c)3=a3+b3+c3+3a2(b+c)+b2(c+a)+c2(a+b)+6abc a3+b3+c3+6abc≥a2(b+c)+b2(c+a)+c2(a+b) Recalling that Schur’s inequality gives a3+b3+c3+3abc ≥ a2(b+c)+b2(c+a)+c2(a+b), the inequality follows. In particular, equality necessitates that the extra 3abc on the left is 0. Combined with the equality condition of Schur, we have equality where two of a, b, c are 1 and the third is 0. This is a typical dumbass solution. 2 Solution 2. Without loss of generality, take a ≥ b ≥ c. As a+b+c=1, we 1+c≤3 or 1?3c≥ 0. Write the left hand side as (a+b)3?3ab(a + b ? 2c) = (a + b)3 ? 3ab(1 ? 3c). This is minimized for a ?xed sum a + b where ab is made as large as possible. As by AM-GM (a + b)2 ≥ 4ab, this minimum occurs if and only if a = b. 3 2 Hence, we need only consider the one variable inequality 2 1?c + c3 + 6 1?c c = 2 2 1 · (9c3 ? 9c2 + 3c + 1) ≥ 1 . Since c ≤ 1 , 3c ≥ 9c2 . Dropping this term and 9c3 , the 4 4 3 inequality follows. Particularly, 9c3 = 0 if and only if c = 0, and the equality cases are 1 when two variables are 2 and the third is 0. This solution is of the smoothing variety in that we moved two variables together while preserving their sum. In other inequalities √ we may wish to preserve products and thus analyze the assignment a = b = ab. These are both representative of the technique of mixing variables, in which two or more variables are blended together algebraically as part of an argument allowing us to outright equate variables


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