Proof that circuit upper bounds for $E$ imply $P \neq NP$ In the official Clay problem description for P versus NP it's stated that $P \neq NP$ would follow from showing that "every language in $E$ [the class of languages recognizable in exponential time with a deterministic Turing machine] can be computed by a Boolean circuit family $&lt;B_n&gt;$ such that for at least one $n$, $B_n$ has fewer gates than the maximum needed to
compute any Boolean function $f: \{0,1\}^n \longrightarrow \{0,1\}$." However, the only reference is that this "is an intriguing observation by V. Kabanets." Could someone please point me to a published version of this implication with the proof?
I don't think the paper in the other answer contains an answer to your question. Indeed I am not really sure a proof has been published, because the result follows from other well known results. 

The proof of the statement you want is as follows: 


$\Sigma_3 E$ contains a function of maximum possible circuit complexity on every input length, by simply defining a function which proves itself (using alternation) to be different from all functions with non-maximum circuit complexity. This is standard and the proof idea can be found in sources such as Arora and Barak's textbook.
If $P = NP$ then $\Sigma_3 E = E$, by padding and the collapse of the polynomial time hierarchy to $P $.
Therefore if $P = NP$ then there is a language in $E$ with maximum circuit complexity. This is the contrapositive of what you want to prove.