Mathematics OpenStudy (anonymous):

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 $$\frac 32$$ OpenStudy (anonymous):

And yes we don't need to use Lagrange multiplier. ( http://en.wikipedia.org/wiki/Lagrange_multiplier ) otherwise it will be boring. OpenStudy (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:$\sqrt{1-\frac{1}{x^2+1}}+\sqrt{1-\frac{1}{y^2+1}}+\sqrt{1-\frac{1}{z^2+1}}$ OpenStudy (asnaseer):

which means we need to maximize each of the denominators OpenStudy (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 OpenStudy (asnaseer):

but I cannot prove my /gut/ is correct :) OpenStudy (anonymous):

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

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

No. OpenStudy (asnaseer):

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

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

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

$x \sqrt{y ^{2}+1}\sqrt{z ^{2}+1} = x \sqrt{(y ^{2}+1)(z ^{2}+1})$ OpenStudy (anonymous):

$x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}$ OpenStudy (anonymous):

Your a liar, This is hard OpenStudy (blockcolder):

Oh, jensen's. Therefore, asnaseer's approach is worth looking at. OpenStudy (anonymous):

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

If you know jensen's this is just there lines. OpenStudy (asnaseer):

never had the pleasure of meeting him :) OpenStudy (anonymous):

yeah you guys can just leave me do this alone OpenStudy (anonymous):

You were here at 1906 ? :P OpenStudy (asnaseer):

of course - weren't you? OpenStudy (asnaseer):

19:06 == 6 minutes past 7pm OpenStudy (anonymous):

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

Haha :D OpenStudy (anonymous):

$2({\sqrt{3}-y}) + 2x(\sqrt{3} - x -y)$ OpenStudy (anonymous):

$2\sqrt{3} - 2y + 2{\sqrt{3}}x - 2x ^{2} - 2xy$ OpenStudy (anonymous):

differentiate to find max value OpenStudy (anonymous):

I'm exhausted OpenStudy (blockcolder):

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

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

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

*of OpenStudy (anonymous):

I was successful in avoiding the Hessian by keeping things only to one variable. OpenStudy (asnaseer):

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

so then we need to maximize:$\frac{x'}{\sqrt{x'^2+3}}+...$ OpenStudy (asnaseer):

sorry, that should be - we need to maximize:$\frac{x'\sqrt{3}}{\sqrt{3x'^2+1}}+...$ OpenStudy (asnaseer):

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

Latest Questions vdancy100: which sentence best describes the excerpt from shakeshears sonnet 130
17 minutes ago 0 Replies 0 Medals ag209893: hi ?
3 hours ago 23 Replies 0 Medals JeliseWh05: 6y+20c-3x=
1 hour ago 1 Reply 0 Medals 71611: Karen currently has a account balance
1 hour ago 1 Reply 0 Medals ilovemyfaimaly: i think this is right and correct me if i am wrong but... if the equation seid what is : 9/4 of 35/2 = "Of" is the same as multiplication so, 9/4 u00d7 35/2
1 hour ago 2 Replies 0 Medals ilovemyfaimaly: i think this is right and correct me if i am wrong but... if the equation seid what is : 9/4 of 35/2 = "Of" is the same as multiplication so, 9/4 u00d7 35/2
6 hours ago 1 Reply 0 Medals ilovemyfaimaly: i think this is right and correct me if i am wrong but... if the equation seid what is : 9/4 of 35/2 = "Of" is the same as multiplication so, 9/4 u00d7 35/2
6 hours ago 0 Replies 0 Medals sammyc1313: You have figured out the marginal cost and marginal benefit of buying an extra smoothie.
7 hours ago 0 Replies 0 Medals alpha: osmosis and diffusion help ?
3 hours ago 6 Replies 0 Medals austin24523: Which is an example of something that amuses Mr. Laurence about Jo? her honesty a
7 hours ago 5 Replies 0 Medals