Math Question and Answers


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 189269 by Rupesh123 last updated on 14/Mar/23

	Answered by a.lgnaoui last updated on 15/Mar/23
	$posons:E=(1/a)+(1/b)+(1/c)+((15)/(a+b+c)) (a+b+c)×E=15+(a+b+c)((1/a)+(1/b)+(1/c)) =18+((a(b+c))/(bc))+((b(a+c))/(ac))+((c(a+b))/(ab)) =18+(((1−2abc−ab)c+ab(a+b))/(abc)) =18+(((1−2abc−ab)/(ab)) +((a+b)/c)) (i) 2abc+(a+b)c+ab=1 (a>0,b>0,c>0) 1−2abc−ab=(a+b)c { ((E(a+b+c)=18+(((a+b)c)/(ab))+((a+b)/c))),((E=((18)/((a+b+c)))+(([(a+b+c)−c)]c)/(ab(a+b+c)))+(((a+b+c)−c)/(c(a+b+c))))) :} posons ( a+b+c)=λ ((18)/λ)+(((λ−c)c)/(λab))+((λ−c)/(λc))=((18)/λ)+((λ−c)/λ)((c/(ab))+(1/c)) =(1/λ)[18+((λ/c)−1)(1+(c^2 /(ab)))] =(1/λ)[17+((((a+b)c)/(ab))+(λ/c) ) =((17abc)/((a+b+c)))+(((a+b)c)/((a+b+c)ab))+(1/c) pour tout a b c>0 abc>(a+b+c) ⇒((abc)/λ)>1 (((a+b)c)/(λab))+(1/c)>0⇒E≥17 ⇒(1/a)+(1/b)+(1/c)+((15)/(a+b+c))≥16 i$
	$p o s o n s : E = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{15}{a + b + c}$ $(a + b + c) \times E = 15 + (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ $= 18 + \frac{a (b + c)}{b c} + \frac{b (a + c)}{a c} + \frac{c (a + b)}{a b}$ $= 18 + \frac{(1 - 2 a b c - a b) c + a b (a + b)}{a b c}$ $= 18 + (\frac{1 - 2 a b c - a b}{a b} + \frac{a + b}{c}) (i)$ $2 a b c + (a + b) c + a b = 1 (a > 0, b > 0, c > 0)$ $1 - 2 a b c - a b = (a + b) c$ ${\begin{cases} E (a + b + c) = 18 + \frac{(a + b) c}{a b} + \frac{a + b}{c} \\ E = \frac{18}{(a + b + c)} + \frac{[(a + b + c) - c)] c}{a b (a + b + c)} + \frac{(a + b + c) - c}{c (a + b + c)} \end{cases}$ $p o s o n s (a + b + c) = λ$ $\frac{18}{λ} + \frac{(λ - c) c}{λ a b} + \frac{λ - c}{λ c} = \frac{18}{λ} + \frac{λ - c}{λ} (\frac{c}{a b} + \frac{1}{c})$ $= \frac{1}{λ} [18 + (\frac{λ}{c} - 1) (1 + \frac{c^{2}}{a b})]$ $= \frac{1}{λ} [17 + (\frac{(a + b) c}{a b} + \frac{λ}{c})$ $= \frac{17 a b c}{(a + b + c)} + \frac{(a + b) c}{(a + b + c) a b} + \frac{1}{c}$ $p o u r t o u t a b c > 0 a b c > (a + b + c)$ $\Rightarrow \frac{a b c}{λ} > 1 \frac{(a + b) c}{λ a b} + \frac{1}{c} > 0 \Rightarrow E ⩾ 17$ $\Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{15}{a + b + c} ⩾ 16$ $i$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum