Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 188387 by cortano12 last updated on 28/Feb/23

Answered by Frix last updated on 28/Feb/23

$$\mathrm{I}\mathrm{think}\mathrm{because}\mathrm{of}\mathrm{symmetry}a=b=c=1\Rightarrow$$$$\mathrm{Minimum}\mathrm{is}\frac{3}{2}$$

Answered by mnjuly1970 last updated on 28/Feb/23

$$$$$$E=\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{ab+bc}+\frac{\frac{1}{c^2}}{ac+bc}$$$$\stackrel{T_{2-}lemma}{⩾}\frac{{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^2}{2(ab+ac+bc)}$$$$\stackrel{abc=1}{=}\frac{{(ab+ac+bc)}^2}{2(ab+ac+bc)}$$$$=ab+ac+bc\stackrel{am-gm}{⩾}\frac{3}{2}\sqrt[3]{a^2{b}^2{c}^2}$$$$min\left(E\right)=\frac{3}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com