if x;y;z>0 and xyz=8 prove that: (1/(x^2 + 8)) + (1/(y^2 + 8)) + (1/(z^2 + 8)) ≤ (1/4)


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 149036 by mathdanisur last updated on 02/Aug/21

$i f x; y; z > 0 a n d x y z = 8 p r o v e t h a t :$ $\frac{1}{x^{2} + 8} + \frac{1}{y^{2} + 8} + \frac{1}{z^{2} + 8} ⩽ \frac{1}{4}$
	Answered by dumitrel last updated on 02/Aug/21

	$p^{3} ⩾ 27 r \Rightarrow p ⩾ 6$ $q^{2} ⩾ 3 p r \Rightarrow q ⩾ 12$ $Σ \frac{1}{x^{2} + 8} = \frac{3}{8} + Σ (\frac{1}{x^{2} + 8} - \frac{1}{8}) = \frac{3}{8} - \frac{1}{8} Σ \frac{x^{2}}{x^{2} + 8} ⩽$ $⩽ \frac{3}{8} - \frac{1}{8} \cdot \frac{p^{2}}{p^{2} - 2 q + 24} \overset{∙}{⩽} \frac{1}{4}$ $∙ \Leftrightarrow \frac{p^{2}}{p^{2} - 2 q + 24} ⩾ 1 \Leftrightarrow q ⩾ 12$
	Commented by mathdanisur last updated on 02/Aug/21

	$C o o l, T h a n k Y o u S e r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum