74th Romanian Mathematical Olympiad

The given relation is equivalent to: $${\left(1+\frac{b}{a}\right)}^x-b{\left(\frac{1}{a}\right)}^x\ge 1,\forall x\ge \alpha .$$ Considering the function $f:\mathbb{R}\to \mathbb{R}$, $f(x)=(1+\frac{b}{a}{)}^x-b(\frac{1}{a}{)}^x$, we can notice that the function is increasing, being a sum of two increasing functions. Therefore, because $f(1)=1$, we obtain that $f(x)\ge f(1)=1$, for each $x\ge 1$, while $f(x)<f(1)=1$, for each $x<1$. Therefore, the minimum value of $\alpha$ is 1.