nãy là cách 1:
Cách 2: Từ \(a^2+b^2+c^2=1\)
\(\Rightarrow F=ab+bc+2ac=a^2+b^2+c^2+ab+bc+2ca-1\)
\(=\left[\left(a^2+2ac+c^2\right)+b\left(a+c\right)+\frac{b^2}{4}\right]+3\cdot\frac{b^2}{4}-1\)
\(=\left[\left(a+c\right)^2+b\left(a+c\right)+\frac{b^2}{4}\right]+3\cdot\frac{b^2}{4}-1\)
\(=\left[a+c+\frac{b}{2}\right]^2+3\cdot\frac{b^2}{4}-1\ge-1\)
Cách 3:Ta có: \(\left(a+b+c\right)^2\ge0\left(1\right)\)
\(\left(a+c\right)^2\ge0\left(2\right)\) và \(b^2\ge0\left(3\right)\)
Cộng theo vế (1),(2),(3) có:
\(\left(a+b+c\right)^2+\left(a+c\right)^2+b^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)+a^2+2ac+c^2+b^2\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)+2\left(ab+bc+2ca\right)\ge0\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)+\left(ab+bc+2ca\right)\ge0\)
\(\Leftrightarrow\left(ab+bc+2ca\right)\ge-\left(a^2+b^2+c^2\right)=-1\)
\(\Rightarrow F\ge-1\)