Just another super easy problem,
Let \(x, y, z \in \mathbb{R}^+\) such that \(x +y +z =\sqrt{3} \) .
Prove that the maximum value of \(\large \frac x{\sqrt{x^2+1}} +\frac y{\sqrt{y^2+1}} +\frac z{\sqrt{z^2+1}}\) is \(\frac 32 \)

- anonymous

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- jamiebookeater

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- anonymous

And yes we don't need to use Lagrange multiplier. ( http://en.wikipedia.org/wiki/Lagrange_multiplier ) otherwise it will be boring.

- asnaseer

we could write:\[\frac{x}{\sqrt{x^2+1}}=\sqrt{\frac{x^2}{x^2+1}}=\sqrt{\frac{x^2+1-1}{x^2+1}}=\sqrt{1-\frac{1}{x^2+1}}\]so then we want to maximize:\[\sqrt{1-\frac{1}{x^2+1}}+\sqrt{1-\frac{1}{y^2+1}}+\sqrt{1-\frac{1}{z^2+1}}\]

- asnaseer

which means we need to maximize each of the denominators

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## More answers

- asnaseer

which gives me the /gut/ feeling that x=y=z is the solution since we have a constraint on the sum of x, y and z

- asnaseer

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

- anonymous

The reason why I liked this problem is the fact that it uses an inequality which is very useful in general.

- asnaseer

is it worth pursuing:
let x = a\(\sqrt{3}\)
let y = b\(\sqrt{3}\)
let z = c\(\sqrt{3}\)
so we have: a + b + c = 1

- blockcolder

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

- anonymous

No.

- asnaseer

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Your a liar, This is hard

- blockcolder

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

- anonymous

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

- anonymous

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

- asnaseer

never had the pleasure of meeting him :)

- anonymous

yeah you guys can just leave me do this alone

- anonymous

You were here at 1906 ? :P

- asnaseer

of course - weren't you?

- asnaseer

19:06 == 6 minutes past 7pm

- anonymous

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

- anonymous

Haha :D

- anonymous

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

- anonymous

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

- anonymous

differentiate to find max value

- anonymous

I'm exhausted

- blockcolder

Jensen's goes like this, right?
\[\lambda_1+\lambda_2+\cdots+\lambda_n=1\\
f''(x)>0\\
\therefore f(\lambda_1x_1+\lambda_2x_2+\cdots+\lambda_nx_n)\leq\lambda_1f(x_1)+\lambda_2f(x_2)+\cdots+\lambda_nf(x_n) \]

- asnaseer

I /think/ you are correct blockcolder from what I've just read about this method, but I don't understand it well enough to apply it here.

- anonymous

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

- asnaseer

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

- asnaseer

*of

- anonymous

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

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

- asnaseer

I'm probably way off the mark as its very late here and I need to get some sleep.
good like with this guys.
I'll check-in sometime tomorrow (or should I say later today) to how how this was solved.

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