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

$a \leq 1$

$a + b \leq 5$

$a + b + c \leq 14$

$a + b + c + d \leq 30$

Prove: $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} \leq 10$

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 $x_{1}$ , ..., $x_{n}$ , $y_{1}$ , ..., $y_{n}$ are non-negative and:

$x_{1} \leq y_{1}$

$x_{1} + x_{2} \leq y_{1} + y_{2}$

...

$x_{1} + \dots + x_{n} \leq y_{1} + \dots + y_{n}$

then

$\sqrt{x_{1}} + \dots + \sqrt{x_{n}} \leq \sqrt{y_{1}} + \dots + \sqrt{y_{n}}$

Then the result follows by setting $n = 4$ , $x_{1} = a$ , $x_{2} = b$ , $x_{3} = c$ , $x_{4} = d$ , $y_{1} = 1$ , $y_{2} = 4$ , $y_{3} = 9$ , $y_{4} = 16$ .

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 $A = \sqrt{a} - 1$

Let $B = \sqrt{b} - 2$

Let $C = \sqrt{c} - 3$

Let $D = \sqrt{d} - 4$

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

$a = A^{2} + 2 A + 1$

$b = B^{2} + 4 B + 4$

$c = C^{2} + 6 C + 9$

$d = D^{2} + 8 D + 16$

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

$a = A^{2} + 2 A + 1 \leq 1$

$a + b = A^{2} + B^{2} + 2 A + 4 B + 5 \leq 5$

$a + b + c = A^{2} + B^{2} + C^{2} + 2 A + 4 B + 6 C + 14 \leq 14$

$a + b + c + d = A^{2} + B^{2} + C^{2} + D^{2} + 2 A + 4 B + 6 C + 8 D + 30 \leq 30$

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

$A^{2} + 2 A \leq 0$

$A^{2} + B^{2} + 2 A + 4 B \leq 0$

$A^{2} + B^{2} + C^{2} + 2 A + 4 B + 6 C \leq 0$

$A^{2} + B^{2} + C^{2} + D^{2} + 2 A + 4 B + 6 C + 8 D \leq 0$

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

$2 A \leq 0$

$2 A + 4 B \leq 0$

$2 A + 4 B + 6 C \leq 0$

$2 A + 4 B + 6 C + 8 D \leq 0$

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

$6 (2 A) \leq 0$

$2 (2 A + 4 B) \leq 0$

$1 (2 A + 4 B + 6 C) \leq 0$

$3 (2 A + 4 B + 6 C + 8 D) \leq 0$

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

$24 A + 24 B + 24 C + 24 D \leq 0$

$A + B + C + D \leq 0$

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

$\sqrt{a} - 1 + \sqrt{b} - 2 + \sqrt{c} - 3 + \sqrt{d} - 4 \leq 0$

$\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} \leq 10$

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

y_{i}

increasing. It is very easy to find a counterexample otherwise (e.g.

a \leq 4

a + b \leq 5

, with

a = 2.25

b = 2.56

, gives

\sqrt{a} + \sqrt{b} = 3.1 > \sqrt{4} + \sqrt{1}

). The proof is very similar to the solution of the problem.

Kerwin