ta sử dụng bđt :\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)(dk mọi abcd)
cái này cm dễ thôi. bunhia nha
ĐĂT :\(A=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)
\(\Rightarrow A=\sqrt{\left(x+\frac{y}{2}\right)^2+\left(\frac{y\sqrt{3}}{2}\right)^2}+\sqrt{\left(y+\frac{z}{2}\right)^2+\left(\frac{z\sqrt{3}}{2}\right)^2}+\sqrt{\left(z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}\right)^2}\)
Áp dingj bđt trên ta được \(A\ge\sqrt{\left(x+\frac{y}{2}+y+\frac{z}{2}+z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}+\frac{y\sqrt{3}}{2}+\frac{z\sqrt{3}}{2}\right)^2}\)
\(\Rightarrow A\ge\sqrt{\frac{9}{4}\left(x+y+z\right)^2+\frac{3}{4}\left(x+y+z\right)^2}=\sqrt{3}\left(x+y+z\right)\)(dpcm)
Dấu = xảy ra khi và chỉ khi x=y=z
\(\sqrt{x^2+xy+y^2}=\sqrt{\frac{3}{4}\left(x+y\right)^2+\frac{1}{4}\left(x-y\right)^2}\ge\sqrt{\frac{3}{4}\left(x+y\right)^2}=\frac{\sqrt{3}}{2}\left(x+y\right)\)
cách khác
ÁP DỤNG BĐT Mincopxki
\(VT=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\)
\(=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}}+\sqrt{\left(y+\frac{z}{2}\right)^2+\frac{3z^2}{4}}+\sqrt{\left(x+\frac{z}{2}\right)^2+\frac{3z^2}{4}}\)
\(\ge\sqrt{\left(x+y+z+\frac{x+y+z}{2}\right)^2+\left(\frac{\sqrt{3}\left(x+y+z\right)}{2}\right)^2}\)
\(=\sqrt{\frac{9\left(x+y+z\right)^2}{4}+\frac{3\left(x+y+z\right)^2}{4}}\)
\(=\sqrt{3\left(x+y+z\right)^2}=\sqrt{3}\left(x+y+z\right)=VP\)
