Let a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0 Prove a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0, where n is any positive number

Let a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0 Prove a ^ n * (a ^ 2-B * c) + B ^ n (b ^ 2-ac) + C ^ n (C ^ 2-AB) > = 0, where n is any positive number

The original inequality is equivalent to a ^ (n + 2) + B ^ (n + 2) + C ^ (n + 2) ≥ a ^ NBC + B ^ NAC + C ^ nab. Let a ≤ B ≤ C, then ab ≤ AC ≤ BC. According to the sorting inequality: A ^ NBC + B ^ NAC + C ^ nab (sum of reverse order) ≤ a ^ nab + B ^ NBC + C ^ NAC = a ^ (n + 1) B + B ^ (n + 1) C + C ^ (n + 1) a (sum of disorder order) ≤ a ^ (n + 1) a + B ^