## zepdrix one year ago Let a, b, and c be non-negative real numbers. Prove the following inequality:

1. zzr0ck3r

same here

2. zepdrix

$$\large\rm ab+bc+ca\le a^2+b^2+c^2$$ So here is was I was attempting... For $$\large\rm a,~b,~c\ge0$$ we have $$\large\rm (a+b+c)^2\ge0$$. Expanding this out gives us: $$\large\rm a^2+b^2+c^2+2ab+2bc+2ca\ge0$$

3. misty1212

idk but i think you are going to need something else, since you did not really use the fact that they are non negative

4. misty1212

$(x+b+c)^2\geq 0$ no matter what

5. zepdrix

Oh haha good point XD

6. dan815

how about thinking about it like the sum sub of 3 perfect square vs the sum of interchanging perfect squares

7. dan815

so like rectangles formed from mixing and matching the sides of perfect squares

8. dan815

|dw:1441246380649:dw|

9. zzr0ck3r

Use Cauchy / Shwarz

10. misty1212

i think the trick is to write $a^2+b^2+c^2-ab-bc-ab$ as a sum of squares

11. dan815

|dw:1441246436682:dw|

12. zepdrix

Doh >.< Yah teacher mentioned something about Cauchy, I thought it was for a different problem though.. hmm

13. zzr0ck3r

You are allowed?

14. misty1212

think we can do it without that maybe

15. dan815

hey but loook my picture works!!

16. zepdrix

I am allowed to use Cauchy Schwarz if I first first prove it lol. That's a separate problem though I suppose :)

17. zepdrix

The squares thing? :O works?

18. dan815

if you arrange any set of A^2,B^2,C^2 whjere A>B>C

19. zzr0ck3r

it is not hard to prove for $$\mathbb{R}$$.

20. misty1212

$a^2+b^2+c^2-ab-ac-bc=(a-b)^2+(b-c)^2+(c-a)^2$ or maybe half of that

21. zzr0ck3r

Else, I would assume wlog a<=b<=c to get rid of so many cases.

22. misty1212

since the right hand side is clearly positive, you get $a^2+b^2+c^2-ab-ac-bc\geq 0$

23. anonymous

you don't need cases, do what @misty1212 said, except the algebra is not quite right

24. zepdrix

Oh so the idea I started with was kind of a mess? Where do you come up with the subtraction like that though? :d hmm

25. anonymous

it is a common gimmick is all i think

26. zepdrix

Ooo that works out very nicely Misty!! :O

27. zepdrix

Hmm I have a couple more like this :) Imma give them a shot though before I bug you guys.

28. zzr0ck3r

Another great trick in these is to add 0. like +a-a ...

29. zepdrix

ya ya ya ya ya :) good call boss

30. zepdrix

I'll post the next one just in case anyone wanted to look at it :d$\large\rm 8abc\le(a+b)(b+c)(c+a)$

31. zepdrix

hmm

32. zzr0ck3r

Well, if you expand the RHS you get 8 terms, maybe use the last one?

33. ganeshie8

suppose x,y,z are positive real numbers can we say min(xyz) = min(x)*min(y)*min(z) ?

34. zepdrix

ya you get 8 terms :d two of which combine to give us some abc's. Hmm. min what? +_+

35. ganeshie8

$\min\{(a+b)(b+c)(c+a)\} =\min\{a+b)\} *\min\{b+c\}* \min\{c+a\}$ By AM-GM inequality we have $$a+b\ge 2\sqrt{ab}$$ giving the desired result

36. ganeshie8

presuming $$a,b,c$$ are positive reals which keeps the sums $$a+b, ~b+c, ~c+a$$ also positive

37. zepdrix

Hmm neato :) Thanks