$(a^{3}+b^3)(b^3+c^3)(c^3+a^3)=a^3b^3c^3$
`<=>` $(a+b)(a^{2}-ab+b^2)(b+c)(b^2-bc+c^2)(a+c)(c^2-ac+a^2)=a^3b^3c^3$
`<=>` $(a+b)(b+c)(a+c)(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)=a^3b^3c^3$
`<=>` $abc(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)=a^3b^3c^3$
`<=>` $abc(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^3b^3c^3=0$
`<=>` $abc\left[\begin{array}{ccc}(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^2b^2c^2\\\end{array}\right]=0$
`<=>` \(\left[ \begin{array}{l}abc=0(1)\\\left[\begin{array}{ccc}(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^2b^2c^2\\\end{array}\right]=0(2)\end{array} \right.\)
`<=>` `(``2``)` $(a^{2}-ab+b^2)=a^2-2a.\dfrac{b}{2}+\dfrac{b^2}{4}+\dfrac{3b^2}{4}=(a-\dfrac{b}{2})^2+\dfrac{3b^2}{4}$
`Ta` `có :` $(a-\dfrac{b}{2})^{2}\ge0(\forall a,b)$
`=>` $(a-\dfrac{b}{2})^{2}+\dfrac{3b^2}{4}\ge\dfrac{3b^2}{4}$
$\text{Tương tự ta có:}$
$(b-\dfrac{c}{2})^{2}+\dfrac{3c^2}{4}\ge\dfrac{3c^2}{4}$ $(c-\dfrac{a}{2})^{2}+\dfrac{3a^2}{4}\ge\dfrac{3a^2}{4}$
`Vậy` $(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)^{}=\dfrac{27}{64}a^2b^2c^2(3)$
$\text{Từ (2) và(3) suy ra:}$
`=>` $a^{2}b^2c^2=0$
`=>` $abc=0(4)$ $\text{(1)(4) => abc=0 (đpcm)}$
`@K`
