\(5x^2+xy+5y^2=\dfrac{9}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{9}{4}\left(x+y\right)^2+\dfrac{1}{2}\left(x+y\right)^2=\dfrac{11}{4}\left(x+y\right)^2\)
... từ đó suy ra
\(A\ge\dfrac{\sqrt{11}}{2}\left(x+y\right)+\dfrac{\sqrt{11}}{2}\left(y+z\right)+\dfrac{\sqrt{11}}{2}\left(z+x\right)=\sqrt{11}\left(x+y+z\right)=a\sqrt{11}\)
\(5x^2+xy+y^2=\dfrac{11}{4}\left(x+y\right)^2+\dfrac{9}{4}\left(x-y\right)^2\ge\dfrac{11}{4}\left(x+y\right)^2\)
chứng minh tương tự
\(5y^2+yz+5z^2\ge\dfrac{11}{4}\left(y+z\right)^2\)
\(5z^2+xz+5x^2\ge\dfrac{11}{4}\left(x+z\right)^2\)
=> \(A\ge\dfrac{\sqrt{11}}{2}\left(x+y+y+z+x+z\right)=\sqrt{11}\left(x+y+z\right)=a\sqrt{11}\)
dấu "=" xảy ra khi x=y=z=a/3
