Question

Cho x;y;z >0 thỏa mãn x+ y + z ≤ 1. Chứng minh rằng :

\(17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge35\)

Answer 1

Áp dụng BĐT BSC và BĐT Cosi:

\(17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\ge17\left(x+y+z\right)+\dfrac{2.\left(1+1+1\right)^2}{x+y+z}\)

\(=17\left(x+y+z\right)+\dfrac{18}{x+y+z}\)

\(=17\left(x+y+z\right)+\dfrac{17}{x+y+z}+\dfrac{1}{x+y+z}\)

\(\ge2\sqrt{17\left(x+y+z\right).\dfrac{17}{x+y+z}}+\dfrac{1}{1}\)

\(=35\)

\(\Rightarrow17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge35\)

Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}\)

Answer 2

\(17x+\dfrac{17}{9x}\ge\dfrac{34}{3}\)

tương tự.....

suy ra 

\(17\left(x+y+z\right)+\dfrac{17}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{34}{3}.3=34\)

lại có 

\(\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{9}{x+y+z}.\dfrac{1}{9}=1\)

nên

\(17\left(x+y+z\right)+\dfrac{17}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=17\left(x+y+z\right)+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge35\)