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Iranian Mathematical Olympiad

Algebra Difficulty 5.9 AIME, harder Prove it Iran

Let a1,…,an;b1,…,bn;c1,…,cna_1, \dots, a_n; b_1, \dots, b_n; c_1, \dots, c_n be real numbers. Prove that
∑i=1n(3ai−bi−ci)2+∑i=1n(3bi−ai−ci)2+∑i=1n(3ci−ai−bi)2≥∑i=1nai2+∑i=1nbi2+∑i=1nci2.(→p.21) \sqrt{\sum_{i=1}^{n} (3a_i - b_i - c_i)^2} + \sqrt{\sum_{i=1}^{n} (3b_i - a_i - c_i)^2} + \sqrt{\sum_{i=1}^{n} (3c_i - a_i - b_i)^2} \\ \geq \sqrt{\sum_{i=1}^{n} a_i^2} + \sqrt{\sum_{i=1}^{n} b_i^2} + \sqrt{\sum_{i=1}^{n} c_i^2}. \qquad (\to \text{p.21})

Solution

According to the Minkowski's inequality we have
∑i=1n(3ai−bi−ci)2+∑i=1nbi2+∑i=1nci2≥∑i=1n(3ai−bi−ci+bi+ci)2=3∑i=1nai2. \sqrt{\sum_{i=1}^{n} (3a_i - b_i - c_i)^2} + \sqrt{\sum_{i=1}^{n} b_i^2} + \sqrt{\sum_{i=1}^{n} c_i^2} \\ \geq \sqrt{\sum_{i=1}^{n} (3a_i - b_i - c_i + b_i + c_i)^2} = 3\sqrt{\sum_{i=1}^{n} a_i^2}.
Adding the above inequality with other two similar inequalities, give us the desired result. ■

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