Given that $\frac{1}{x}>(y+z)$, where x<0

Multiplying 'x' both sides, (inequality sign to be flipped since x is negative)

$x*\frac{1}{x}>x(y+z)$

Or, $1 < xy+xz$

Ans. (D)


QS: can we write the reciprocal as -$\frac{1}{x}$ and then proceed. please correct me if my thought process is wrong. YES, IT'S WRONG.

Suppose x = -5, reciprocal of x is $\frac{1}{x}$ = $\frac{1}{(-5)}$ = -$\frac{1}{5}$; If you say, reciprocal of x is -$\frac{1}{x}$, then -$\frac{1}{x}$ = $\frac{-1}{-5}$ = $\frac{ 1}{5}$.