Question

Cho 3 số dương x,y,z thoả mãn x+y+z=1. Chứng minh:
\(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}>14\)

Answer 1

\(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{3}{x^2+y^2+z^2}\)

\(\ge\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{x^2+y^2+z^2+2xy+yz+xz}=\frac{\left(\sqrt{6}+\sqrt{3}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{3}\right)^2\)

(*) ta CM :\(\left(\sqrt{6}+\sqrt{3}\right)^2>14\)

TA có \(\left(\sqrt{6}+\sqrt{3}\right)^{^2}=6+3+2\sqrt{18}=9+6\sqrt{2}>9+5=14\)

=> \(\frac{3}{xy+yz+xz}+\frac{3}{x^2+y^2+z^2}>14\)