Chiều Max:∏(a+b)≤(2(a+b+c)3)3=64
dấu bằng xảy ra khi a=b=c=2
Chiều Min:
Do a,b,c∈[1;3] nên:
.∏(a−1)≥0⇔abc−∑ab+∑a−1≥0
⇔abc+5≥∑ab
.∏(a−3)≤0⇔abc−3∑ab+9∑a−27≤0
⇔abc+27≤3∑ab
Do đó, abc+27≤3(abc+5)⇒abc≥6(1)⇒∑ab≥11(2)
Mà ∏(a+b)=(∑ab)(∑a)−abc=6∑ab−abc(3)
Since (1),(2) and (3), where are problems in my solution?