Ta có: $a^2\ge a^2-(b-c)^2=(a+b-c)(a-b+c)=(1-2c)(1-2b)$
Tương tự: $b^2\ge(1-2a)(1-2c);c^2\ge(1-2a)(1-2b)$
Từ đó suy ra: $abc\ge(1-2a)(1-2b)(1-2c)$
$\Leftrightarrow abc\ge1-2(a+b+c)+4(ab+bc+ca)-8abc$
$\Leftrightarrow ab+bc+ca\le\dfrac{1+9abc}{4}$
$\Leftrightarrow ab+bc+ca-2abc\le\dfrac{1+abc}{4}$
Lại có: $abc\le\dfrac{(a+b+c)^3}{27}=\dfrac{1}{27}$
Suy ra: $ab+bc+ca-2abc\le\dfrac{1+\dfrac{1}{27}}{4}=\dfrac{7}{27}$
Dấu bằng xảy ra khi: $a=b=c=\dfrac{1}{3}$