Solution 28 Dec 2006

If $a,b,c,x\,$ are real numbers such that $a,b,c>0\,$ and

$\frac{xb + (1-x)c}{a} = \frac{xc + (1-x)a}{b} = \frac{xa + (1-x) b }{c} \qquad\qquad(1)$

then prove that $a = b = c\,$ .

Proof. Assume that $a,b,c\,$ are not all equal.

Note that equation (1) has a cyclic symmetry, i.e. it is unchanged if we replace $a,b,c\,$ by $b,c,a\,$ (or $c,a,b\,$ ) respectively.

Take the first two terms of (1) and multiply by $ab\,$ , then isolate $x\,$ :

$x(b^2-bc-ac+a^2) = a^2 - bc\qquad\qquad(2)$

Likewise, by cyclic symmetry, we have

$x(c^2-ca-ba+b^2) = b^2 - ca\,$

$x(a^2-ab-cb+c^2) = c^2 - ab\,$

Adding these three equations yields

$2 x (a^2+b^2+c^2-ab-bc-ca) = a^2+b^2+c^2-ab-bc-ca\,$

By the rearrangement inequality, we have that $a^2+b^2+c^2\,>\, ab+bc+ca$ (this uses that $a,b,c\,$ are not all equal), so $x=\frac12$ .

Then (2) becomes

$(b-a)(a+b+c) = 0\,$

But $a+b+c>0\,$ because $a,b,c>0\,$ , so this implies $b=a\,$ . By symmetry this implies $a=b=c\,$ , contradicting the assumption.

$\square$