Non-uniform lower bounds from uniform It is well known that lower bounds on circuits implies lower bounds on time complexity, just because circuits can efficiently simulate algorithms. 

By the continuous work started by Ryan Williams, we know that upper bounds on the algorithms can imply circuit lower bounds [1].

Another connection between uniform and non-uniform complexities is that there exist algorithms that uses efficient circuits to gain a speed up. For example, first subquadratic algorithm for 3-sum [2] uses efficient decision trees. 

I'm looking for results in the direction I haven't discussed yet: can lower bounds for time complexity imply lower bounds for circuit complexity. It is worth mentioning that algorithms that uses efficient circuits are taking advantage of the actual constructions rather then pure existence.

[1] Improving exhaustive search implies superpolynomial lower bounds. Ryan Williams.

[2]Threesomes, Degenerates, and Love Triangles. Allan Grønlund, Seth Pettie
MA $\subseteq$ PSPACE $\subseteq$ EXP $\subseteq$ NEXP ​ and ​ P/poly has undecidable unary languages, so

NEXP $\subseteq$ P/poly ​ $\implies$ ​ NEXP $\subset$ P/poly ​ $\implies$ ​ NEXP $\subseteq$ EXP .

As something like a generalization of that ​ (showing that EXP-complete languages are

in CHECK is difficult) , ​ for all languages L in CHECK with a checker whose queries all have

length at most ​ nd , ​ for all non-decreasing functions ​ S : {0,1,2,3,4,...} $\to$ {0,1,2,3,4,...} :

If L's circuit-size in bits is at most S then

L ​ $\in$ ​ DSPACE$\left(\hspace{-0.04 in}\left(\hspace{-0.02 in}n^d\hspace{-0.05 in}\cdot \hspace{-0.03 in}S\hspace{-0.05 in}\left(\hspace{-0.03 in}n^d\hspace{-0.02 in}\right)\hspace{-0.04 in}\right)\hspace{-0.04 in}+\hspace{-0.03 in}poly\hspace{.02 in}(n)\hspace{-0.03 in}\right)$ ​ $\subseteq$ ​ DTIME$\hspace{-0.02 in}\left(\hspace{-0.05 in}2^{\left(n^{\hspace{.02 in}d} \hspace{-0.02 in}\cdot S\left(\hspace{-0.02 in}n^{\hspace{.02 in}d}\hspace{-0.03 in}\right)\hspace{-0.03 in}\right)+\hspace{.02 in}poly\hspace{.02 in}(n)}\hspace{-0.04 in}\right)$ ​ ​ ,

so by contrapositive, if ​ ​ ​ L ​ $\not\in$ ​ DTIME$\hspace{-0.02 in}\left(\hspace{-0.05 in}2^{\left(n^{\hspace{.02 in}d} \hspace{-0.02 in}\cdot S\left(\hspace{-0.02 in}n^{\hspace{.02 in}d}\hspace{-0.03 in}\right)\hspace{-0.03 in}\right)+\hspace{.02 in}poly\hspace{.02 in}(n)}\hspace{-0.04 in}\right)$

then L's circuit-size in bits exceeds S infinitely-often.

("in bits" as as opposed to "in gates".)

Similarly, for all functions T and $\ell$ and $S$ from {0,1,2,3,...} to {1,2,3,4,5,...},

for all search problems R in ​ FNTIME(T) ​ whose output-lengths are bounded above by $\ell$:

If R's circuit-size in bits is at most S then R is solvable by ​ FDTIME$\left(\hspace{-0.02 in}2^{S(n)} \hspace{-0.04 in}\cdot \hspace{-0.04 in}poly\hspace{.02 in}(S(n))\hspace{-0.04 in}\cdot \hspace{-0.04 in}T(n)\hspace{-0.03 in}\right)$ ,

so by contrapositive, if R is not solvable by ​ FDTIME$\left(\hspace{-0.02 in}2^{S(n)} \hspace{-0.03 in}\cdot \hspace{-0.03 in}\left(\left(\hspace{.05 in}poly\hspace{.02 in}(S(n)) \cdot \ell(n)\right)+ T(n)\right)\hspace{-0.03 in}\right)$

then R's circuit-size in bits exceeds S infinitely-often.