# If a + b + c = abc, prove that at least one of (a,b,c) is < or = sqrt(3)

1. Jun 5, 2004

### maverick280857

Hi

Here's a question....

Let a, b, c be three non-zero real numbers such that

a + b + c = abc

Prove that at least one of these three numbers (a, b or c) is less than or equal to the square root of 3.

Can you prove this without trigonometry? The trigonometric solution follows...

Solution (using Trigonometry)

Let a = tan(A), b = tan(B), c = tan(C) (for some nonzero angles A, B, C which are real) so that the given constraint becomes

tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C)

which can be true iff A + B + C = n*pie (n is an integer)

If n = 1, then A, B, C are the angles of a triangle (as the constraint is true for angles of a triangle). The result follows by considering cases: of an equilateral triangle where A = B = C = pie/3 radians so that each of a, b and c is equal to sqrt(3); next consider the case of a scalene triangle where A, B and C are all distinct. If A > pie/3, then B+C = pie-A = pie-(qty less than pie/3) and so either B or C is less than pie/3.

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Cheers
Vivek

2. Jun 5, 2004

### arildno

i) $$a=b=c=\sqrt{3}$$ is a solution of the equation
ii) Setting $$a=\sqrt{3}+\hat{a}$$ and similarly for b and c

The proof follows readily from this
(i.e., at least one of the hatted numbers must be non-positive.)

Last edited: Jun 5, 2004
3. Jun 6, 2004

### maverick280857

Hi

Thanks. Your method is interesting...we get a new constraint on the hatted numbers now...am I right? (I haven't completed the solution using your substitutions..)

Cheers
Vivek

4. Jun 6, 2004

### matt grime

proof without using trig or substitution: use the arithmetic mean/geometric mean inequality

5. Jun 6, 2004

### Gokul43201

Staff Emeritus
Yes, you end up with an equation which can be written like

$$f(\hat{a},\hat{b},\hat{c}) = -g(\hat{a},\hat{b},\hat{c})$$

And you'll find that for positive values of $$\hat{a},\hat{b},\hat{c}$$, the functions f and g must give positive numbers. So you have LHS = positive number and RHS = negative number...a contradiction ! So, one of $$\hat{a},\hat{b},\hat{c}$$ must not be positive.

Last edited: Jun 6, 2004
6. Jun 6, 2004

### maverick280857

Thanks Gokul43201

I get it now :-D

Cheers
Vivek

7. Jun 6, 2004

I don't =[