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if-x-y-z-gt-0-and-xyz-8-prove-that-1-x-2-8-1-y-2-8-1-z-2-8-1-4-




Question Number 149036 by mathdanisur last updated on 02/Aug/21
if   x;y;z>0   and   xyz=8   prove that:  (1/(x^2  + 8)) + (1/(y^2  + 8)) + (1/(z^2  + 8)) ≤ (1/4)
$${if}\:\:\:{x};{y};{z}>\mathrm{0}\:\:\:{and}\:\:\:{xyz}=\mathrm{8}\:\:\:{prove}\:{that}: \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\:\mathrm{8}}\:+\:\frac{\mathrm{1}}{{y}^{\mathrm{2}} \:+\:\mathrm{8}}\:+\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} \:+\:\mathrm{8}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Answered by dumitrel last updated on 02/Aug/21
p^3 ≥27r⇒p≥6  q^2 ≥3pr⇒q≥12  Σ(1/(x^2 +8))=(3/8)+Σ((1/(x^2 +8))−(1/8))=(3/8)−(1/8)Σ(x^2 /(x^2 +8))≤  ≤(3/8)−(1/8)∙(p^2 /(p^2 −2q+24))≤^• (1/4)  •⇔(p^2 /(p^2 −2q+24))≥1⇔q≥12
$${p}^{\mathrm{3}} \geqslant\mathrm{27}{r}\Rightarrow{p}\geqslant\mathrm{6} \\ $$$${q}^{\mathrm{2}} \geqslant\mathrm{3}{pr}\Rightarrow{q}\geqslant\mathrm{12} \\ $$$$\Sigma\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{8}}+\Sigma\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\Sigma\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{8}}\leqslant \\ $$$$\leqslant\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{{p}^{\mathrm{2}} }{{p}^{\mathrm{2}} −\mathrm{2}{q}+\mathrm{24}}\overset{\bullet} {\leqslant}\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\bullet\Leftrightarrow\frac{{p}^{\mathrm{2}} }{{p}^{\mathrm{2}} −\mathrm{2}{q}+\mathrm{24}}\geqslant\mathrm{1}\Leftrightarrow{q}\geqslant\mathrm{12} \\ $$
Commented by mathdanisur last updated on 02/Aug/21
Cool, Thank You Ser
$${Cool},\:{Thank}\:{You}\:{Ser} \\ $$

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