if-x-y-z-gt-1-then-x-1-y-1-z-1-x-1-y-1-z-1-lt-xyz-8-

Question Number 147892 by mathdanisur last updated on 24/Jul/21

i f x; y; z > 1 t h e n :

\sqrt{\frac{(x - 1) (y - 1) (z - 1)}{(x + 1) (y + 1) (z + 1)}} < \frac{x y z}{8}

Answered by mindispower last updated on 24/Jul/21

\frac{x - 1}{x + 1} < \frac{x^{2}}{4} \dots ? \forall x > 1

\Leftrightarrow 4 (x - 1) < x^{2} (x + 1)

\Leftrightarrow x^{3} + x^{2} + 4 - 4 x > 0

\Leftrightarrow x (x^{2} + 1) + 4 - 4 x > 0

x^{2} + 1 ⩾ 2 x \Rightarrow x (x^{2} + 1) + 4 - 4 x > 2 x^{2} - 4 x + 4 = 2 {(x - 1)}^{2} + 2 > 0

\Rightarrow \forall x > 1 \frac{x - 1}{x + 1} < \frac{x^{2}}{4}, \frac{y - 1}{y + 1} < \frac{y^{2}}{4}, \frac{z - 1}{z + 1} < \frac{z^{2}}{4}

\Rightarrow \frac{(x - 1) (y - 1) (z - 1)}{(x + 1) (y + 1) (z + 1)} < \frac{x^{2} y^{2} z^{2}}{64}

\Rightarrow \sqrt{\frac{(x - 1) (y - 1) (z - 1)}{(x + 1) (y + 1) (z + 1}} < \frac{x y z}{8}

Commented by mathdanisur last updated on 24/Jul/21

T h a n k y o u S i r c o o l