I'd like to elaborate on some of the fine points that Arthur raised.
The $\aleph$ numbers (also the $\beth$ numbers) are used to denote cardinals. However one of the key features of cardinals is that we can say "the next cardinal", and we can say which cardinal came first and which came second. These are ordinal properties.
Note that the least cardinal greater than $\aleph_{\aleph_0}$ also has countably many [infinite] cardinals smaller than itself. But since $\aleph_0+1=\aleph_0$, what sense would that make?
So we are using the ordinals. It's a fine point, because the finite notions coincide, the finite cardinals are the finite ordinals, and it's not until we reach the infinite ordinals that we run into the difference between $\omega$ and $\aleph_0$.
Therefore, instead of $\aleph_{\aleph_0}$ we have $\aleph_\omega$, then we have $\aleph_{\omega+1}$ and so on and so forth. After we have gone through uncountably many of these we finally have $\aleph_{\omega_1}$, where $\omega_1$ is the least uncountable ordinal -- which corresponds to $\aleph_1$.
And so on and so forth. For every ordinal $\alpha$ we have $\aleph_\alpha$ is the unique cardinal that the infinite cardinals below it have the same order type as $\alpha$.
The point here is what does cardinal mean.
In many contexts, and perhaps naively, one might consider cardinals to mean initial ordinals, so that a set which cannot be well-ordered does not have a cardinal. In this context it is not hard to prove that $\kappa+\kappa=\kappa$ for any infinite cardinal, simply because this easily holds for well-ordered cardinals.
In many contexts, which are different, and also perhaps naively, one considers a cardinal to be something representing cardinality of sets. This can be done in multiple ways, and it doesn't really matter, since cardinal arithmetic is really just the "quotient" of considering sets and injections, along with unions and products. In this case, you will need the axiom of choice. Suppose, for example, that $A$ is a set such that if $B\subseteq A$ then $B$ is finite or $A\setminus B$ is finite. These sort of sets are called amorphous and their existence is consistent with $\sf ZF$.
But now consider $|A|+|A|$, it's the cardinal of $A\times\{0,1\}$, which certainly can be split into two infinite subsets, $A\times\{0\}$ and $A\times\{1\}$. So, $|A|\neq|A|+|A|$ in that case. There are other examples that are consistent.
Sageev proved that if $\kappa+\kappa=\kappa$ for any infinite cardinal, in the latter sense, then we can deduce that every infinite set has a countably infinite subset (which, for example, prevents the existence of amorphous sets). But at the same time, we cannot even prove that every countable family of non-empty sets admits a choice function. So while the axiom of choice is needed, its "absolute necessity" here is very mild and hard to notice.
Finally, if we do assume the axiom of choice, then every set can be well-ordered, so the two contexts merge into one, and then we can prove the equality in a straightforward manner by replacing a set with whatever ordinal is in bijection with it.
Best Answer
If you can compare $b$ and $c$, then suppose without loss of generality that $b\leq c$. Then $$bc \leq cc = c \leq b+c \leq c+c = 2c \leq bc.$$
The last inequality because we are assuming $b$ and $c$ are both infinite, so $2\leq b$; the equality $cc=c$ by assumption.
Added${}^{\mathbf{2}}$. From $a^2=a$ for all infinite cardinals, one may deduce the Axiom of Choice (this is a theorem of Tarski's), which in turn is equivalent to the fact that we can compare $b$ and $c$.
If you don't want to go through AC in order to assume comparability of $b$ and $c$, then I would have had some trouble with the problem, though Apostolos's answer together with your $b+c = b+bc+c\geq bc$ does tie it together neatly.
Added. In ZF, the Axiom of Choice is equivalent to the statement that given any two sets $A$ and $B$, either $A$ injects into $B$ or $B$ injects into $A$ (that is, the cardinalities of $A$ and $B$ are comparable). Whether or not we can assume that two cardinals are comparable may depend on what one means by "cardinal".
By "cardinal", I understood a cardinal number, which means an ordinal that is not bijectable with any strictly smaller ordinal, where ordinals are ordered by $\in$ (this is the definition in Jech's Set Theory, under Alephs, page 24: "An ordinal $\alpha$ is called a cardinal number if $|\alpha|\neq|\beta|$ for all $\beta\lt\alpha$." The definition precedes the discussion of the Axiom of Choice, which begins in page 38). The ordering among cardinals is induced by the ordering of ordinals. If that is what we mean by "cardinal", then any two cardinals are certainly comparable (even in ZF without AC), so we will either have $b\leq c$ or $c\leq b$, and there is no loss in generality in assuming the first. But as Asaf points out, when dealing with cardinal numbers in this sense, both $a^2=a$ and $b+c=bc=\max\{b,c\}$ are theorems in ZF.