Let k = (((xy + yz + zx)(x + y + z))/((x + y)(y + z)(z + x))) Find the minimum and maximum value of k .


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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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