Áp dụng BĐT Mincopxki ta có:
\(M=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\)
\(=\sqrt{\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}}+\sqrt{\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}}+\sqrt{\left(z+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4}}\)
\(\ge\sqrt{\left(x+y+z+\dfrac{x+y+z}{2}\right)^2+\left(\dfrac{\sqrt{3}\left(x+y+z\right)}{2}\right)^2}\)
\(\ge\sqrt{\left(1+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Lời giải:
Ta có: \(x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\)
Mà \((x-y)^2\geq 0\forall x,y\in\mathbb{R}\Rightarrow x^2+xy+y^2\geq \frac{3}{4}(x+y)^2\)
\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}}{2}|x+y|\)
Tương tự:
\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}|y+z|; \sqrt{z^2+zx+x^2}\geq \frac{\sqrt{3}}{2}|x+z|\)
Cộng các BĐT trên thu được:
\(M\geq \frac{\sqrt{3}}{2}(|x+y|+|y+z|+|z+x|)\geq \frac{\sqrt{3}}{2}|2x+2y+2z|\)
\(\Leftrightarrow M\geq \frac{\sqrt{3}}{2}.2=\sqrt{3}\)
Vậy \(M_{\min}=\sqrt{3}\Leftrightarrow x=y=z=\frac{1}{3}\)
