Đặt \(\left(x;2y;4z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2019\)
Ta cần chứng minh: \(\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{c+a}\le2019\)
Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow VT=\frac{2ab}{a+b}+\frac{2bc}{b+c}+\frac{2ac}{c+a}\le\frac{2ab}{4}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{2bc}{4}\left(\frac{1}{b}+\frac{1}{c}\right)+\frac{2ac}{4}\left(\frac{1}{c}+\frac{1}{a}\right)\)
\(\Rightarrow VT\le\frac{b}{2}+\frac{a}{2}+\frac{c}{2}+\frac{b}{2}+\frac{a}{2}+\frac{c}{2}=a+b+c=2019\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{2019}{3}\) hay \(\left(x;y;z\right)=\left(\frac{2019}{3};\frac{2019}{6};\frac{2019}{12}\right)\)
