## FoolForMath Group Title 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 $$\frac 32$$ 2 years ago 2 years ago

1. FoolForMath Group Title

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

2. asnaseer Group Title

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:$\sqrt{1-\frac{1}{x^2+1}}+\sqrt{1-\frac{1}{y^2+1}}+\sqrt{1-\frac{1}{z^2+1}}$

3. asnaseer Group Title

which means we need to maximize each of the denominators

4. asnaseer Group Title

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

5. asnaseer Group Title

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

6. FoolForMath Group Title

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

7. asnaseer Group Title

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

8. blockcolder Group Title

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

9. FoolForMath Group Title

No.

10. asnaseer Group Title

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

11. FoolForMath Group Title

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

12. FoolForMath Group Title

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

13. ninhi5 Group Title

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

14. ninhi5 Group Title

$x \sqrt{y ^{2}+y ^{2}z ^{2}+z ^{2}+y ^{2}z ^{2}}$

15. luisrock2008 Group Title

Your a liar, This is hard

16. blockcolder Group Title

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

17. ninhi5 Group Title

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

18. FoolForMath Group Title

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

19. asnaseer Group Title

never had the pleasure of meeting him :)

20. ninhi5 Group Title

yeah you guys can just leave me do this alone

21. FoolForMath Group Title

You were here at 1906 ? :P

22. asnaseer Group Title

of course - weren't you?

23. asnaseer Group Title

19:06 == 6 minutes past 7pm

24. ninhi5 Group Title

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

25. FoolForMath Group Title

Haha :D

26. ninhi5 Group Title

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

27. ninhi5 Group Title

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

28. ninhi5 Group Title

differentiate to find max value

29. ninhi5 Group Title

I'm exhausted

30. blockcolder Group Title

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

31. asnaseer Group Title

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.

32. FoolForMath Group Title

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

33. asnaseer Group Title

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

34. asnaseer Group Title

*of

35. FoolForMath Group Title

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

36. asnaseer Group Title

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

37. asnaseer Group Title

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

38. asnaseer Group Title

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

39. asnaseer Group Title

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

40. asnaseer Group Title

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.