Question

The minimum value of
$\sqrt{x {}^{2} + \frac{1}{y {}^{2} } } + \sqrt{ y {}^{2} + \frac{1}{z {}^{2} } } + \sqrt{z {}^{2} + \frac{1}{x {}^{2} } } = a$
Find
$a^{2}$

Answer 1

Mark brainelist bro please
$\sqrt{ {x}^{2} + \frac{1}{ {y}^{2} } } + \sqrt{ {y}^{2} + \frac{1}{ {z}^{2} } } + \sqrt{ {z}^{2} + \frac{1}{ {x}^{2} } } = a \\ on \: squaring \: both \: sides \: we \: have \\ = > {a}^{2} = ( { \sqrt{ {x}^{2} + \frac{1}{ {y}^{2} } } + \sqrt{ {y}^{2} + \frac{1}{ {z}^{2} } } + \sqrt{ {z}^{2} + \frac{1}{ {x}^{2} } }})^{2} \\ = > {a}^{2} = x + \frac{1}{y} + y + \frac{1}{z} + z + \frac{1}{x} + 2 \\ = > {a}^{2} = \frac{ {x}^{2} yz + xz + x {y}^{2} z + xy + xy {z}^{2} + yz + 2xyz }{xyz}$