Ta có:
\(\sqrt{2019x+yz}=\sqrt{x\left(x+y+z\right)+yz}\)\(=\sqrt{x^2+xy+xz+yz}=\sqrt{x^2+yz+x\left(y+z\right)}\)
Áp dụng BĐT AM-GM cho các số không âm, ta có:
\(x^2+yz\ge2x\sqrt{yz}\)
\(\Rightarrow x^2+yz+x\left(y+z\right)\ge x\left(y+z+2\sqrt{yz}\right)\)
\(\Leftrightarrow2019x+yz\ge\left[\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\right]^2\)\(\ge0\)
\(\Rightarrow\sqrt{2019x+yz}\ge\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow x+\sqrt{2019x+yz}\ge\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow\frac{x}{x+\sqrt{2019x+yz}}\le\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
CMTT, ta có:
\(\frac{y}{y+\sqrt{2019y+zx}}\le\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\),\(\frac{z}{z+\sqrt{2019z+xy}}\le\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\Rightarrow M\le\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
\(''=''\Leftrightarrow x=y=z=673\)
Đề là \(M=\sum\frac{x}{x+\sqrt{2019+yz}}\) hay \(M=\sum\frac{x}{x+\sqrt{2019x+yz}}\) bạn?
Nếu là đề bạn đúng thì mình bó tay.
