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 125868 by I want to learn more last updated on 14/Dec/20

Answered by mindispower last updated on 14/Dec/20

(1/(2x))+(1/(2y))≥(2/(x+y)).⇔(x+y)^2 ≥4xy,x,y>0 ⇔(x−y)^2 ≥0 true ⇒∀(x,y)∈R_+ ^2 (1/(2x))+(1/(2y))≥(2/(x+y)) (1/a)+(1/b)+(1/c)=(1/(2a))+(1/(2b))+(1/(2a))+(1/(2c))+(1/(2b))+(1/(2c))...E (1/(2a))+(1/(2b))≥(2/(a+b)) (1/(2a))+(1/(2c))≥(2/(a+c)),(1/(2c))+(1/(2b))≥(2/(c+b)) E⇒(1/a)+(1/b)+(1/c)≥(2/(a+b))+(2/(a+c))+(2/(b+c))

12x+12y⩾2x+y.⇔(x+y)2⩾4xy,x,y>0⇔(x−y)2⩾0true⇒∀(x,y)∈R+212x+12y⩾2x+y1a+1b+1c=12a+12b+12a+12c+12b+12c...E12a+12b⩾2a+b12a+12c⩾2a+c,12c+12b⩾2c+bE⇒1a+1b+1c⩾2a+b+2a+c+2b+c

Commented by I want to learn more last updated on 15/Dec/20

Thanks sir.

Thankssir.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com