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

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

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


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

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

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

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

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

[QED]