Ta có: \(x^2+49y^2\ge14xy\) ; \(x^2+49z^2\ge14zx\) ; \(7y^2+7z^2\ge14yz\)
Cộng vế với vế:
\(2x^2+56y^2+56z^2\ge14\left(xy+yz+zx\right)=14\)
\(\Leftrightarrow x^2+28y^2+28z^2\ge7\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{7}{\sqrt{15}};\frac{1}{\sqrt{15}};\frac{1}{\sqrt{15}}\right)\)