Theo BĐT tam giác ta có:
\(b+c>a\Rightarrow a+b+c>2a\Rightarrow2>2a\Rightarrow a< 1\)
Tương tự cũng có: \(b<1;c<1\)
Áp dụng BĐT AM-GM ta có:
\(\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(\dfrac{1-a+1-b+1-c}{3}\right)^3=\left(\dfrac{3-\left(a+b+c\right)}{3}\right)^3=\dfrac{1}{27}\)
\(\Rightarrow0< \left(1-a\right)\left(1-b\right)\left(1-c\right)\le\dfrac{1}{27}\)
\(\Rightarrow0< ab+bc+ca-abc-\left(a+b+c\right)+1\le\dfrac{1}{27}\)
\(\Rightarrow0< ab+bc+ca-abc-1\le\dfrac{1}{27}\)
\(\Rightarrow1< ab+bc+ca-abc\le\dfrac{28}{27}\)
\(\Rightarrow2< 2ab+2bc+2ca+a^2+b^2+c^2-\left(a^2+b^2+c^2+2abc\right)\le\dfrac{56}{27}\)
\(\Rightarrow2< \left(a+b+c\right)^2-\left(a^2+b^2+c^2+2abc\right)\le\dfrac{56}{27}\)
\(\Rightarrow2< 4-\left(a^2+b^2+c^2+2abc\right)\le\dfrac{56}{27}\)
\(\Rightarrow\dfrac{52}{27}\le a^2+b^2+c^2+2abc< 2\) *Đúng*
