Let Q>0\mathbb{Q}_{>0} be the set of positive rational numbers. Let f:Q>0Rf: \mathbb{Q}_{>0} \to \mathbb{R} be a function satisfying the following three conditions:

(i) for all x,yQ>0x, y \in \mathbb{Q}_{>0}, we have f(x)f(y)f(xy)f(x)f(y) \geq f(xy);

(ii) for all x,yQ>0x, y \in \mathbb{Q}_{>0}, we have f(x+y)f(x)+f(y)f(x + y) \geq f(x) + f(y);

(iii) there exists a rational number a>1a > 1 such that f(a)=af(a) = a.

Prove that f(x)=xf(x) = x for all xQ>0x \in \mathbb{Q}_{>0}.