(a+b)³ = a³ + 3a²b+ 3ab² + b³
Different equivalent forms:
(1)：(a+b)³ = a³ + b³+3ab(a+b)
(2)：a³ + b³ = (a+b)³ – 3ab(a+b)
(3): a³ + b³ = (a+b)(a² -ab + b²)
(4)：(a+b)³ – ( a³ + b³ ) = 3ab(a+b)
1997 USAMO Q5:
Prove:
Proof:
Apply (3):
a³ + b³ = (a+b)(a² -ab + b²) ≥ (a+b)ab
Note:
a² -ab + b²= (a-b)² + ab ≥ ab
since (a-b)² ≥ 0
Symmetrically,
Add 3 RHS:
[QED]
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