Say $X$ is a smooth algebraic variety, $U$ is a Zariski open set in $X$, $L$ is a local system on $U$, and $IC(L)$ the intersection cohomology sheaf on $X$ which restricts to $L$ on $U$. Then is: 


$$\dim L\_u \ge \sum\_i \dim \mathrm{H}^i(\mathrm{IC}(L)\_x) $$ 


for $u \in U$ and all $x \in X$?


If not, is it true after e.g. putting in some scalar depending only on the dimension of $X$? 



Bounding the size of stalks of IC sheaves
The answer to the first question is certainly "no": you can easily find example of X whose IC complex (for the trivial local system of rank 1) has (total) stalk of dimension $\geq 1$ (for instance this is true for most nilpotent orbit closures in a simple Lie algebra).


Edit: Sorry I missed the assumption that X is smooth. But still I think the following is a counterexample.
First, let $Y$ in ${\mathbb C}^n$ be a generic homogeneous hypersurface of degree d (it is smooth away from 0).
Let $Z$be its projectivization. Then if I am not mistaken, the stalks of the IC sheaf of $Y$ at $0$ live in 
dimensions $-(n-1)$ and $-(n-2)$ and they are equal to $H^0(Z)$ and $H^1(Z)$ respectively (I am using perverse normalization). 
Now take $n=3$. Then $Z$ is a curve of degree $d$ in $\mathbb P^2$, so its genus is $g=\frac{(d-1)(d-2)}{2}$
and its $H^1$ has dimension $2g$. Now there is a finite map $\pi:Y\to \mathbb C^2$ of degree $d$ (take for example
$Y$ to be given by the equation $x^d+y^d+z^d=0$ and consider the projection to $(x,y)$).
Then $\pi\_\*$ of the constant sheaf is going to be equal to its Goresky-Macpherson extension from an open subset, where it will be equal to a local system $L$ of rank $d$. But the sum on the right hand side of your expression for $x=0$ is
$1+(d-1)(d-2)$. 


