Let a, B, C ∈ [0,2], and prove that 4A + B ^ 2 + C ^ 2 + ABC ≥ 2Ab + 2BC + 2ca

a、b、c∈[0,2]
So a (b-2) (C-2) > = 0
The expansion is a (bc-2b-2c + 4) > = 0
4a+abc>=2ac+2bc
B ^ 2 + C ^ 2 > = 2BC
The sum is 4A + B ^ 2 + C ^ 2 + ABC ≥ 2Ab + 2BC + 2ca

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