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experimentX
 4 years ago
Let \( a, b, c > 0 \), show that
\[ \sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3. \]
from
http://math.stackexchange.com/questions/193568/inequalitysqrtfrac11a5a6bsqrtfrac11b5b6csqrtfrac11c
experimentX
 4 years ago
Let \( a, b, c > 0 \), show that \[ \sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3. \] from http://math.stackexchange.com/questions/193568/inequalitysqrtfrac11a5a6bsqrtfrac11b5b6csqrtfrac11c

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TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yeah this is gonna be a pain with LaTeX down :S

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0isn't latex working for you? this Q seems fairly complicated ... I've been stalking for more than 2 hrs.

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1no, GoDaddy isn't supporting mathjax right now, there is an announcement in the yellow bar on the home page, Why, is it working for you o.O ?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I'm using chrome ... what server does MathJax uses to parse latex?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1not sure, not that tech savvy... but I have chrome as well. do you see the TeX normally or not?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0I see it normally ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let us suppose a is the largest number out of the the three...... then dw:1347343904345:dw will have the largest value.... So now, this value is maximum when a approaches infinity So,dw:1347343979969:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If a approaches infinity then surely dw:1347344106656:dw will be zero

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And the term dw:1347344150915:dw also cannot be greater than 1.48

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0as a, or b, or c approaches infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The maximum value is when a=b=c

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0let a>=b>=c.. now 11a / 5a + 6b = 11/ 5 + 6(b/a) > whole term always less than or equal toone since b>=a thus sqrt (of this) <= 1 for 11b/5b + 6c = 11/5 + 6(c/b) > same thing 11c/5c + 6a = 11(c/a) /5(c/a) + 6 let c/a = m => 11m / 5m + 6 = 11/5 ( m / m + 6/5) = 2.2 ( m + 6/5  6/5) /(m+6/5) =2.2 ( 1  1.2/(m+1.2) ) note that m>=1 for m tends to infinity , max value becomes 2.2 and sqrt(2.2) is approx 1.4 something.. other 2 terms were less than 1 and this one's maxima is 1.4 something,, hence there sum is always less than 3..and equality holds as a=b=c..
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