Given a, B, C ∈ R +, prove: A ^ 12 / BC + B ^ 12 / Ca + C ^ 12 / AB > = a ^ 10 + B ^ 10 + C ^ 10, use sorting inequality to solve

Given a, B, C ∈ R +, prove: A ^ 12 / BC + B ^ 12 / Ca + C ^ 12 / AB > = a ^ 10 + B ^ 10 + C ^ 10, use sorting inequality to solve

Let a > = b > = C, so (a ^ 12 / BC + B ^ 12 / Ca + C ^ 12 / AB) = (a ^ 10 * a ^ 2 / BC + B ^ 10 * B ^ 2 / AC + C ^ 10 * C ^ 2 / AB), then (a ^ 10 * a ^ 2 / BC + B ^ 10 * B ^ 2 / AC + C ^ 10 * C ^ 2 / AB) > = 1 / 3 * (a ^ 10 + B ^ 10 + C ^ 10) * (a ^ 2 / BC + B ^ 2 / AC + C ^ 2 / AB)
And (a ^ 2 / BC + B ^ 2 / AC + C ^ 2 / AB) gets > = 3 from the mean inequality,
So (a ^ 12 / BC + B ^ 12 / Ca + C ^ 12 / AB) = (a ^ 10 * a ^ 2 / BC + B ^ 10 * B ^ 2 / AC + C ^ 10 * C ^ 2 / AB) > = 1 / 3 * (a ^ 10 + B ^ 10 + C ^ 10) * (a ^ 2 / BC + B ^ 2 / AC + C ^ 2 / AB) > = a ^ 10 + B ^ 10 + C ^ 10
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