Question

cho x,y,z>0 thỏa mãn 4x+9y+16z=49. CM\(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\ge49\)

Answer 1

Ta có:

\(\left(4x+9y+16z\right)\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(\sqrt{\frac{4x}{x}}+\sqrt{\frac{9y.25}{y}}+\sqrt{\frac{16z.64}{z}}\right)^2\)

\(\Leftrightarrow49\left(\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\right)\ge\left(2+15+32\right)^2\)

\(\Leftrightarrow\frac{1}{x}+\frac{25}{y}+\frac{64}{z}\ge49\)

Dấu = xảy ra tại \(x=\frac{1}{2};y=\frac{5}{3};z=2\)