## ParthKohli Group Title Can anyone help me with this one? one year ago one year ago

1. ParthKohli Group Title

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-ash4/389738_558445504198872_734761968_n.jpg Since I am too lazy to write everything down :-)

2. ParthKohli Group Title

@amistre64 @.Sam. ?

3. KingGeorge Group Title

Don't we also need the requirement that $$x,y,z\ge1$$?

4. ParthKohli Group Title

Yes.

5. ParthKohli Group Title

Oh, and it's not a homework or a test... just found it and wondered how to solve it. If you know the solution, then please solve it and show complete steps.

6. Bad2zBone Group Title

Are they integers?

7. ParthKohli Group Title

They are anything satisfying $$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2$$

8. wmckinely Group Title

Well, @ParthKohli, it seems like you know more about how to solve this than anyone else here...-_-

9. ParthKohli Group Title

@wmckinely I just know the problem: nothing else ._.

10. Alchemista Group Title

What if you treat it as a minimization/maximization problem.

11. ParthKohli Group Title

that's a very good way to look at it.

12. Alchemista Group Title
13. ParthKohli Group Title

But the biggest problem is how we proceed.

14. ParthKohli Group Title

Never mind, lol. I don't understand that mathematics. :-D

15. KingGeorge Group Title

Well, that might be a problem to understanding....

16. Alchemista Group Title

I'd have to actually do the math. But it looks like a simple application of the processed I linked you to. You have an equality constraint and function both over three variables.

17. Alchemista Group Title

Show that the minimum of the right hand of the inequality is greater than the maximum of left hand.

18. Alchemista Group Title

*greater or equal

19. ParthKohli Group Title

That is not what I am able to understand... how?

20. KingGeorge Group Title

Using some wolfram, this method does not seem to be a simple application of lagrange multipliers.

21. Alchemista Group Title

Ah ok

22. Alchemista Group Title

Did you use both constraints when plugging it into wolfram?

23. KingGeorge Group Title

I may have done it a bit wrong though. Let me double check.

24. Alchemista Group Title

Use the Minimize and Maximize routines.

25. asnaseer Group Title

I /think/ you can do this if you use:$\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}$which I believe can be proved.

26. Alchemista Group Title

Lagrange, worked: Minimize[{Sqrt[x^2 + y^2 + z^2], 1/x + 1/y + 1/z == 2, x > 0, y > 0, z > 0}, {x, y, z}] = Maximize[{Sqrt[x - 1] + Sqrt[y - 1] + Sqrt[z - 1], 1/x + 1/y + 1/z == 2, x > 0, y > 0, z > 0 }, {x, y, z}]

27. Alchemista Group Title

I'll give you the link on wolfram alpha in a moment.

28. asnaseer Group Title

if you multiply both sides of your equation by xyz you get:$yz+xz+xy=2xyz$

29. Alchemista Group Title
30. asnaseer Group Title

then note that:$(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)=x^2+y^2+z^2+4xyz$ from above

31. Alchemista Group Title

Anyways, the solution using multivariable calculus is perfectly valid and works.

32. KingGeorge Group Title

The $$\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}$$ was the fact I was looking for. I kept thinking the inequality was pointing the other direction :/

33. asnaseer Group Title

rearranging slightly we get:$(x+y+z)^2=(x-1)^2+2x-1+(y-1)^2+2y-1+(z-1)^2+2z-1+4xyz$$=(x-1)^2+(y-1)^2+(z-1)^2+2x+2y+2z+4xyz-3$

34. asnaseer Group Title

then take square roots of both sides to get:$x+y+z=\sqrt{(x-1)^2+(y-1)^2+(z-1)^2+\text{other terms from above}}$$\ge\sqrt{(x-1)^2}+\sqrt{(y-1)^2}+\sqrt{z-1)^2}$$\ge(x-1)+(y-1)+(z-1)$

35. asnaseer Group Title

then square root again to get:$\sqrt{x+y+z}\ge\sqrt{(x-1)+(y-1)+(z-1)}$$\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$

36. asnaseer Group Title

I /think/ this works

37. KingGeorge Group Title

Looks good to me.

38. help123please. Group Title

I dont understand that problem,Parth. Obviously.

39. asnaseer Group Title

thx @KingGeorge :)

40. help123please. Group Title

Hi,asnasser. Greetings,George.

41. Alchemista Group Title

In any case you will notice that Lagrange finds the local min/max at (1.5, 1.5, 1.5) for those functions subject to the given constraints.

42. experimentX Group Title

how is $\sqrt{a+b}\ge\sqrt{a}+\sqrt{b}$

43. asnaseer Group Title

\begin{align} (\sqrt{a}+\sqrt{b})^2&=a+b+2\sqrt{ab}\\ \therefore a+b&=(\sqrt{a}+\sqrt{b})^2-2\sqrt{ab}\\ &\ge(\sqrt{a}+\sqrt{b})^2\\ \therefore \sqrt{a+b}&\ge\sqrt{a}+\sqrt{b} \end{align}

44. asnaseer Group Title

assuming a and b are both positive

45. asnaseer Group Title

hope that helps

46. asnaseer Group Title

it may not be a /strict/ proof but it is one that I used to convince myself that this relation might be valid.

47. KingGeorge Group Title

Wait...I'm no longer convinced. Let $$a=9$$, $$b=16$$. Then $$\sqrt{a+b}=5$$ and $$\sqrt{a}+\sqrt{b}=3+4=7>5$$.

48. asnaseer Group Title

:(

49. asnaseer Group Title

where is the flaw in my proof for this?

50. experimentX Group Title

there is a slight error $(\sqrt a + \sqrt b)^2 - 2 \sqrt a \sqrt b \leq (\sqrt a + \sqrt b)^2$ proves the opposite

51. asnaseer Group Title

D'oh! thank for clarifying

52. asnaseer Group Title

but does that mean that I have proved the opposite relation to that being asked for?

53. experimentX Group Title

no ... $\sqrt {x-1}+... \ge \sqrt{x-1 + ... } \le \sqrt{x+y+z}$ condition is not strong enough

54. asnaseer Group Title

Ah! I see now - thanks again @experimentX (and @KingGeorge) Guess I'll have to think again on this problem - maybe calculas is the way to go as you guys showed above.

55. Alchemista Group Title

Why is no one satisfied with treating this as an optimization problem?

56. asnaseer Group Title

It's not that I wasn't satisfied - I just wondered if there was a simpler way of arriving at the solution. :)

57. experimentX Group Title

Lagrange definitely gives solution but I am still looking for more elementary proof.

58. KingGeorge Group Title

There's just something more appealing about the elementary proofs.

59. Alchemista Group Title

I suppose so. At the very least it would be more accessible.