Giải thích các bước giải:

Ta có:

$a^3+b^3+b^3\ge 3ab^2$

$a^3+a^3+b^3\ge 3a^2b$

Cộng vế với vế

$\to 3(a^3+b^3)\ge 3ab(a+b)$

$\to a^3+b^3\ge ab(a+b)$

$\to a^3+b^3+abc\ge ab(a+b+c)$

$\to \dfrac1{a^3+b^3+abc}\le\dfrac1{ab(a+b+c)}$

$\to \dfrac1{a^3+b^3+abc}\le\dfrac{c}{abc(a+b+c)}$

Tương tự:

$\dfrac1{b^3+c^3+abc}\le\dfrac{a}{abc(a+b+c)}$

$\dfrac1{c^3+a^3+abc}\le\dfrac{b}{abc(a+b+c)}$

$\to \dfrac1{a^3+b^3+abc}+\dfrac1{b^3+c^3+abc}+\dfrac1{c^3+a^3+abc}\le \dfrac{c}{abc(a+b+c)}+\dfrac{a}{abc(a+b+c)}+\dfrac{b}{abc(a+b+c)}$

$\to \dfrac1{a^3+b^3+abc}+\dfrac1{b^3+c^3+abc}+\dfrac1{c^3+a^3+abc}\le \dfrac1{abc}$

$\to đpcm$