XVII OBM

Let $Y$ be a maximal subset of $X$ in the sense that if one adjoins another element from $X$ in $Y$ then it will contain a subset from $\mathcal{F}$. Define $f:X\setminus Y\to \left(\genfrac{}{}{0ex}{}{Y}{2}\right)$, where $\left(\genfrac{}{}{0ex}{}{Y}{2}\right)$ is the family of 2-subsets from $Y$, with $f(x)=A$ if $A\cup \{x\}$ is one of the sets from $\mathcal{F}$; if there is more than one set, choose any of them. $f$ is an injective function, because if $f(x_1)=f(x_2)=B$ then $B\cup \{x_1\}$ and $B\cup \{x_2\}$ would be in $\mathcal{F}$ and these two sets would have more than one element in their intersection. Thus, by the injective principle, $$|X\setminus Y|\le \left(\genfrac{}{}{0ex}{}{|Y|}{2}\right)⟺n-|Y|\le \frac{|Y|(|Y|-1)}{2}⟺|Y|\ge -\frac{1}{2}+\sqrt{2n+\frac{1}{4}}$$ Let $\lfloor \sqrt{2n}\rfloor =k$. We need to prove that $-\frac{1}{2}+\sqrt{2n+\frac{1}{4}}>k-1$ in order to prove that $|Y|>k-1⟺|Y|\ge k$. But this is only a small computation: $$-\frac{1}{2}+\sqrt{2n+\frac{1}{4}}>k-1⟺2n+\frac{1}{4}>k^2-k+\frac{1}{4}⟺\sqrt{2n}>\sqrt{k^2-k},$$ which is true because $\sqrt{2n}\ge k>\sqrt{k^2-k}$.