If $$c=\frac{16x}{y}+\frac{49y}{x}$$ for some non-zero real numbers x and y, then c cannot take the value

Question

Accepted Answer

− 50 -50 − 50. Let x y b e t \frac{x}{y}\ be\ t y x ​ b e t Therefore, c = 16 t + 49 t c=16t\ +\ \frac{49}{t} c = 16 t + t 49 ​ Applying AM>= GM ( 16 t + 49 t ) 2 ≥ ( 16 t × 49 t ) 1 2 \frac{\left(16t\ +\ \frac{49}{t}ight)}{2}\ge\ \left(16t	imes\frac{49}{t}ight)^{\frac{1}{2}} 2 ( 16 t + t 49 ​ ) ​ ≥ ( 16 t

If $c=\frac{16x}{y}+\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

If $c=\frac{16x}{y}+\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

If c=16xy+49yxc=\frac{16x}{y}+\frac{49y}{x}c=y16x​+x49y​ for some non-zero real numbers x and y, then c cannot take the value

If c=16xy+49yxc=\frac{16x}{y}+\frac{49y}{x}c=y16x​+x49y​ for some non-zero real numbers x and y, then c cannot take the value

If $c=\frac{16x}{y}+\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

If $c=\frac{16x}{y}+\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value