Hi(adsbygoogle = window.adsbygoogle || []).push({});

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.

--------------------------------------------------------------------------

Cheers

Vivek

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**