Finally $a=b=c$ fails

Algebra Level 4

Let $a,b,c$ be positive reals such that $a+b+c=10$ and $ab+bc+ca=25$ . Let $m=\min(ab,bc,ca)$ . Find the largest possible value of $m$ .

If the largest possible value of $m$ is of the form $\dfrac{m}{n}$ where m and n are coprime positive integers, enter your answer as $m+n$ .

1 solution

Renjith Joshua
Mar 11, 2016

WLOG, we can take ab<bc, ab<ca.

Since all 3 are positive, this gives us a<c and b<c respectively.

The extreme situation occurs when a=b<c.

Using the substitution a=b, the three numbers become $a, a, 10-2a$ .

If you substitute these into $ab+bc+ca=25$ , you'll get $3a^{2} -20a + 25 = 0$ . This gives $a = 5,$ $\frac{5}{3}$ .

We'll reject the former, because it leads to c = 0. The latter gives a = b = 5/3 and c=20/3, which gives m = 25/9.