Let $X$ be a smooth projective variety over an algebraically closed field $k$, and suppose $\dim X=3$. Let $D$ be an effective, very ample divisor and let $x\in X$ be a point. Then is it true that for any prime divisor $S\subseteq X$ which properly intersects $D$ (i.e. $S$ isn't contained in $D$) we have $D^2.S\geq(\operatorname{mult}_x D)^2\cdot\operatorname{mult}_x S$?
I can see that if $\dim X=2$, then $D.S\geq \operatorname{mult}_x D\cdot\operatorname{mult}_x S$. But is the above analogue statement also true? If yes, how to prove it?
What I tried is the following: let $\pi:\widetilde{X}\to X$ be the blow-up of $X$ at $x$, and denote by $\widetilde{D}$ resp. $\widetilde{S}$ the strict transforms of $D$ resp. $S$. Also, denote by $E$ the exceptional divisor, and $\alpha=\operatorname{mult}_x D$ and $\mu=\operatorname{mult}_x S$. Then we have $$ D^2.S=(\pi^*D)^2.\pi^*S=(\widetilde{D}+\alpha E)^2.(\widetilde{S}+\mu E)=\widetilde{D}^2.\widetilde{S}+\alpha^2\mu. $$ So the statement would follow if $\widetilde{D}^2.\widetilde{S}\geq 0$, however why should this hold?