Question

IF X,Y AND Z ARE POSITIVE REAL NUMBERS THEN FIND THE MINIMUM VALUE OF (X+Y+Z)$(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

THIS QUESTION IS REALATED TO SEQUENCE AND SERIES SO SOLVE ON THE BASIS ​

Answer 1

Answer:

$\huge{\boxed{\sf{(x + y + z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) _{min} = 9}}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{A.M\geqslant\:H.M}}}$

SO FROM THESE INEQUALITIES WE SOLVE IN THIS PROCESS.

$\frac{x + y + z}{3} \geqslant \frac{3}{( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) } \\ = )(x + y + z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) \geqslant 9 \\ = )(x + y + z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) _{min} = 9$