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$a;b;c$ là 3 cạnh của tam giác \Rightarrow $1-a;1-b;1-c>0$

Có: $(1-a)(1-b)(1-c) \le (\dfrac{3-a-b-c}{3})^3=\dfrac{1}{27}$

\Leftrightarrow $ab+bc+ca-abc \le \dfrac{28}{27}$

\Leftrightarrow $2abc-2(ab+bc+ca) \ge \dfrac{-56}{27}$

\Leftrightarrow $2abc+a^2+b^2+c^2 \ge \dfrac{-56}{27}+4=\dfrac{52}{27}$

Dấu "=" xảy ra \Leftrightarrow $a=b=c=\dfrac{2}{3}$
 
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