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 Math Help > Inequalities > Inequality Methods > Sum of SQRTs > Sum of SQRTs

## Puzzle question:

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

a≤1,
a+b≤5,
a+b+c≤14,
a+b+c+d≤30.

Prove that sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)≤10

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 + 2A + 1
b = B^2 + 4B + 4
c = C^2 + 6C + 9
d = D^2 + 8D + 16

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

a = A^2 + 2A + 1 <= 1
a+b = A^2 + B^2 + 2A + 4B + 5 <= 5
a+b+c = A^2 + B^2 + C^2 + 2A + 4B + 6C + 14 <= 14
a+b+c+d = A^2 + B^2 + C^2 + D^2 + 2A + 4B + 6C + 8D + 30 <= 30

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

A^2 + 2A <= 0
A^2 + B^2 + 2A + 4B <= 0
A^2 + B^2 + C^2 + 2A + 4B + 6C <= 0
A^2 + B^2 + C^2 + D^2 + 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:

sqrt(a)-1+sqrt(b)-2+sqrt(c)-3+sqrt(d)-4 <= 0

sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d) <= 10

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

### Nick Hobson's Proof

Hi Graeme,

I thought you might be interested in another proof of this inequality, which uses Jensen's inequality

f(x) = sqrt(x) is concave on non-negative reals.

Taking n = 4,
weights 1/10, 2/10, 3/10, 4/10, and
x(i) = a, b/4, c/9, d/16, we obtain

(1/10)*sqrt(a) + (2/10)*sqrt(b/4) + (3/10)*sqrt(c/9) + (4/10)*sqrt(d/16) <= sqrt(a/10 + b/20 + c/30 + d/40).

Hence

sqrt(a) + sqrt(b) + sqrt(c) + sqrt(d) <= 10*sqrt((12a + 6b + 4c + 3d)/120).

But

12a + 6b + 4c + 3d = 3(a + b + c + d) + (a + b + c) + 2(a + b) + 6a <= 3*30 + 14 + 2*5 + 6*1 = 120.

The result follows.

Regards,
Nick

### Related pages in this website

Jensen's inequality can be used to solve the "sum of sqrts" puzzle.

The webmaster and author of this Math Help site is Graeme McRae.