Question

$\text{max} \{ a_1,a_2,....,a_n\}<3 \\ \text{Show that} \\ \\ \frac{1}{3-a_1}&plus;\frac{1}{3-a_2}&plus;...&plus;\frac{1}{3-a_n} \ge \frac{n}{3-\frac{1}{n}(a_1&plus;a_2&plus;...&plus;a_n)} \\ \text{using definition of convex.}$

Answer 1

Given that Manqan, az, ..., an} <3 tet, ai carcassap so.. san<3 Therefore, Marfor, ar, - .. any z an het a function fem -0 . - Where 3-2 1,92,- (3-7) xe • , , , , , Norad, please com , so, tew is isme increasing at à. Hence flos ata is convex function or, using o we have f(x) = $x is convex. Because flas is monotonie increasing fune bet 6, = bz = bg = nr. = bn=1 using convexity propaty on fea) with respect to weigh to bi ; 2=1,2, 3, -2 We get the inequality

SS.. È {( ཙམ་) ༤ པ་ (༤) – bi f (xi) ད་ o; བག- A Since, ait pas, az,--and then we have from ཉ- ; ཉུ - 7 Y] I sinee, bizt vi] e, bi = 1 + - - ཎ ་ 4 – 3- ︽ད་ཉིན） =4 ཅ ཤུལ(གནས་ < 3=ནད * ལན * བ aq to-tar ན ། ， ། - +- རྐྱ- ༢༡ +་ ཉིན་ 3- (or+ དུ+ -+ ཆད་ ཏེ། ཉ=ད་ཀྱི་ proved.