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which means we need to maximize each of the denominators

but I cannot prove my /gut/ is correct :)

AMGM? Cauchy-Schwarz? Chebyshev? (These are the only ones I know.)

No.

I guess thats a no to both blockcolder and myself :(

I like the way asnaseer approaches any problem, always starting from scratch :)

I was talking about Jensen's Inequality ( http://en.wikipedia.org/wiki/Jensen's_inequality )

\[x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})\]

\[x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}\]

Your a liar, This is hard

Oh, jensen's.
Therefore, asnaseer's approach is worth looking at.

x(y+z) do the same thing for all of them

If you know jensen's this is just there lines.

never had the pleasure of meeting him :)

yeah you guys can just leave me do this alone

You were here at 1906 ? :P

of course - weren't you?

19:06 == 6 minutes past 7pm

xy + zy + yx + zy + zx + zy
2zx + 2xy + 2zy
2y(x+z) + 2zx
x+z = sqrt 3 - y

Haha :D

\[2({\sqrt{3}-y}) + 2x(\sqrt{3} - x -y)\]

\[2\sqrt{3} - 2y + 2{\sqrt{3}}x - 2x ^{2} - 2xy\]

differentiate to find max value

I'm exhausted

This is actually concave. f''(x)<0

I /think/ the sum pf your lambda's equate equate to: a+ b + c = 1

*of

I was successful in avoiding the Hessian by keeping things only to one variable.

could we divide everything by \(\sqrt{3}\) to get:\[x'+y'+z'=1\]where:\[x'=\frac{x}{\sqrt{3}}\]etc

so then we need to maximize:\[\frac{x'}{\sqrt{x'^2+3}}+...\]

sorry, that should be - we need to maximize:\[\frac{x'\sqrt{3}}{\sqrt{3x'^2+1}}+...\]

so now the x', y' and z' correspond to your lambda's blockcolder