\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=1+2\left(ab+bc+ca\right).\)
\(\Rightarrow A=\left(ab+bc+ca\right)=\frac{1}{2}\left(a+b+c\right)^2-\frac{1}{2}\ge-\frac{1}{2}\)với mọi a,b,c
Vậy A nhỏ nhất bằng -1/2 khi a+b+c =0
Ta có : \((x-\dfrac{1}{3})^2+(y-\dfrac{1}{3})^2+(z-\dfrac{1}{3})^2>=0\)
\(=>x^2+y^2+z^2-\dfrac{2}{3}(x+y+z)+\dfrac{1}{3}\ge0\)
\(=>x^2+y^2+z^2+\dfrac{1}{3}\ge\dfrac{2}{3}(x+y+z)\)
\(=>1+\dfrac{1}{3}=\dfrac{4}{3}\ge\dfrac{2}{3}(x+y+z)\)
\(=>x+y+z\le2\)
Do đó : \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca).\)
\(=>A=(ab+ac+bc)=\dfrac{1}{2}(a+b+c)^2-\dfrac{1}{2}\le\dfrac{1}{2}.2^2-\dfrac{1}{2}=\dfrac{3}{2}\)
Ta có : \((x-\dfrac{1}{\sqrt{3}})^2+(y-\dfrac{1}{\sqrt{3}})^2+(z-\dfrac{1}{\sqrt{3}})^2>=0\)
\(=>x^2+y^2+z^2-\dfrac{2}{\sqrt{3}}(x+y+z)+1\ge0\)
\(=>x^2+y^2+z^2+1\ge\dfrac{2}{\sqrt{3}}(x+y+z)\)
\(=>1+1=2\ge\dfrac{2}{\sqrt{3}}(x+y+z)\)
\(=>x+y+z\le\sqrt{3}\)
Do đó : \((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca).\)
\(=>A=(ab+ac+bc)=\dfrac{1}{2}(a+b+c)^2-\dfrac{1}{2}\le\dfrac{1}{2}.\sqrt{3}^2-\dfrac{1}{2}=\dfrac{2}{2}=1\)
