IMO Technique

(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:
\frac{1}{a^{3}+b^{3}+abc}   +   \frac{1}{b^{3}+c^{3}+abc}   + \frac{1}{c^{3}+a^{3}+abc}   \leq  \frac{1}{abc}

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

\frac{abc}{a^{3}+b^{3}+abc}       \leq \frac{abc}{(a+b)ab + abc}  = \frac{c}{a+b+c}

Symmetrically,
\frac{abc}{b^{3}+c^{3}+abc}       \leq \frac{a}{a+b+c}

\frac{abc}{c^{3}+a^{3}+abc}       \leq \frac{b}{a+b+c}

Add 3 RHS:
\frac{a+b+c}{a+b+c} = 1

\frac{abc}{a^{3}+b^{3}+abc}   +   \frac{abc}{b^{3}+c^{3}+abc}   + \frac{abc}{c^{3}+a^{3}+abc}   \leq 1

\frac{1}{a^{3}+b^{3}+abc}   +   \frac{1}{b^{3}+c^{3}+abc}   + \frac{1}{c^{3}+a^{3}+abc}   \leq  \frac{1}{abc}

[QED]

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