## ParthKohli 2 years ago Can anyone help me with this one?

1. ParthKohli

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

@amistre64 @.Sam. ?

3. KingGeorge

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

4. ParthKohli

Yes.

5. ParthKohli

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.

Are they integers?

7. ParthKohli

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

8. wmckinely

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

9. ParthKohli

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

10. Alchemista

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

11. ParthKohli

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

12. Alchemista
13. ParthKohli

But the biggest problem is how we proceed.

14. ParthKohli

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

15. KingGeorge

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

16. Alchemista

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

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

18. Alchemista

*greater or equal

19. ParthKohli

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

20. KingGeorge

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

21. Alchemista

Ah ok

22. Alchemista

Did you use both constraints when plugging it into wolfram?

23. KingGeorge

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

24. Alchemista

Use the Minimize and Maximize routines.

25. asnaseer

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

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

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

28. asnaseer

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

29. Alchemista
30. asnaseer

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

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

32. KingGeorge

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

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

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

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

I /think/ this works

37. KingGeorge

Looks good to me.

I dont understand that problem,Parth. Obviously.

39. asnaseer

thx @KingGeorge :)

Hi,asnasser. Greetings,George.

41. Alchemista

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

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

43. asnaseer

\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

assuming a and b are both positive

45. asnaseer

hope that helps

46. asnaseer

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

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

:(

49. asnaseer

where is the flaw in my proof for this?

50. experimentX

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

D'oh! thank for clarifying

52. asnaseer

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

53. experimentX

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

54. asnaseer

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

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

56. asnaseer

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

57. experimentX

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

58. KingGeorge

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

59. Alchemista

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