Sum of Square Roots
By Graeme Mcrae on Thursday, January 03, 2002 - 12:02 am:

This question was given to me, which I thought would be fun for people to think about.


For non-negative numbers a, b, c and d,

a £ 1

a+b £ 5

a+b+c £ 14

a+b+c+d £ 30

Prove: Öa+Öb+Öc+Öd £ 10

Please don't answer this if you've seen it (or one like it) before. I would like to leave the question on the board for a day or two so that people who haven't seen it before can try to work on it. Thanks!

By Michael Doré on Thursday, January 03, 2002 - 03:58 pm:

I think we can prove the more general result:

If x1, ..., xn, y1, ..., yn are non-negative and:

x1 £ y1

x1+x2 £ y1+y2

...

x1+¼+xn £ y1+¼+ yn

then


  __
Öx1
 
 +¼+   __
Öxn
 
 £   __
Öy1
 
 +¼+   __
Öyn
 

Then the result follows by setting n=4, x1=a, x2=b, x3=c, x4=d, y1=1, y2=4, y3=9, y4=16.

By Graeme Mcrae on Tuesday, January 08, 2002 - 04:56 am: A good way to solve this puzzle is to make a substitution. If we define new variables represented by capital A, B, C, and D as follows, then the solution is easier:

Let A=Öa-1

Let B=Öb-2

Let C=Öc-3

Let D=Öd-4

Now, solving for a, b, c, and d, we get the following equations.

a=A2+2A+1

b=B2+4B+4

c=C2+6C+9

d=D2+8D+16

Since a £ 1, and a+b £ 5, etc., we get the following four inequalities:

a=A2+2A+1 £ 1

a+b=A2+B2+2A+4B+5 £ 5

a+b+c=A2+B2+C2+2A+4B+6C+14 £ 14

a+b+c+d=A2+B2+C2+D2+2A+4B+6C+8D+30 £ 30

Notice that in each inequality the same constant appears on both sides, so they cancel:

A2+2A £ 0

A2+B2+2A+4B £ 0

A2+B2+C2+2A+4B+6C £ 0

A2+B2+C2+D2+2A+4B+6C+8D £ 0

A square is always non-negative, so it can be subtracted from the "less than" side of an equality:

2A £ 0

2A+4B £ 0

2A+4B+6C £ 0

2A+4B+6C+8D £ 0

Now multiply the first inequality by 6, the second by 2, the third by 1, and the fourth by 3:

6(2A) £ 0

2(2A+4B) £ 0

1(2A+4B+6C) £ 0

3(2A+4B+6C+8D) £ 0

Now add up the four equations, then divide by 24:

24A+24B+24C+24D £ 0

A+B+C+D £ 0

Now substitute the lower-case expressions in place of A, B, C, and D, and then add 10 to both sides:

Öa-1+Öb-2+Öc-3+Öd-4 £ 0

Öa+Öb+Öc+Öd £ 10


That's it! Pretty easy, don't you think?

By Kerwin Hui on Tuesday, January 08, 2002 - 06:07 pm:
Michael's conjecture should also have the condition yi increasing. It is very easy to find a counterexample otherwise (e.g. a £ 4, a+b £ 5, with a=2.25, b=2.56, gives Öa+Öb=3.1 > Ö4+Ö1). The proof is very similar to the solution of the problem.

Kerwin