Let-k-xy-yz-zx-x-y-z-x-y-y-z-z-x-Find-the-minimum-and-maximum-value-of-k-

Question Number 74322 by naka3546 last updated on 22/Nov/19

L e t

k = \frac{(x y + y z + z x) (x + y + z)}{(x + y) (y + z) (z + x)}

F i n d t h e m i n i m u m a n d m a x i m u m v a l u e o f k .

Answered by MJS last updated on 23/Nov/19

k - 1 = \frac{x y z}{(x + y) (x + z) (y + z)}

y = p x \land z = q x

(1) k - 1 = \frac{p q}{(p + 1) (p + q) (q + 1)}

\frac{d}{d q} [\frac{p q}{(p + 1) (p + q) (q + 1)}] = 0

\frac{p (p - q^{2})}{(p + 1) {(p + q)}^{2} {(q + 1)}^{2}} = 0

\Rightarrow q = \pm \sqrt{p}

insert in (1)

k - 1 = \frac{p}{{(1 \pm \sqrt{p})}^{2} (p + 1)}

\frac{d}{d p} [\frac{p}{{(1 \pm \sqrt{p})}^{2} (p + 1)}] = 0

\frac{1 \pm p^{\frac{3}{2}}}{{(1 \mp \sqrt{p})}^{3} {(p + 1)}^{2}} = 0

\Rightarrow p = 1 \land q = 1

(no other real solution)

\Rightarrow x = y = z

\Rightarrow k - 1 = \frac{1}{8}

\Rightarrow k = \frac{9}{8}

which is the absolute maximum

the minimum is - \infty

put q = 1 :

k - 1 = \frac{p}{2 {(p + 1)}^{2}}

\lim_{p \to - 1} (\frac{p}{2 {(p + 1)}^{2}}) = - \infty

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