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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

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yeah this is gonna be a pain with LaTeX down :S
isn't latex working for you? this Q seems fairly complicated ... I've been stalking for more than 2 hrs.
no, 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 ?

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Other answers:

I'm using chrome ... what server does MathJax uses to parse latex?
not sure, not that tech savvy... but I have chrome as well. do you see the TeX normally or not?
I see it normally ...
Let 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|
If a approaches infinity then surely |dw:1347344106656:dw| will be zero
And the term |dw:1347344150915:dw| also cannot be greater than 1.48
as a, or b, or c approaches infinity
It will be undefined
The maximum value is when a=b=c
let 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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