This article is from the Fusion FAQ, by Robert F. Heeter email@example.com with numerous contributions by others.
Basically speaking, the extra neutrons on the D and T nuclei make
them "larger" and less tightly bound, and the result is
that the cross-section for the D-T reaction is the largest.
Also, because they are only singly-charged hydrogen isotopes,
the electrical repulsion between them is relatively small.
So it is relatively easy to throw them at each other, and it
is relatively easy to get them to collide and stick.
Furthermore, the D-T reaction has a relatively high energy yield.
However, the D-T reaction has the disadvantage that it releases
an energetic neutron. Neutrons can be difficult to handle,
because they will "stick" to other nuclei, causing them to
(frequently) become radioactive, or causing new reactions.
Neutron-management is therefore a big problem with the
D-T fuel cycle. (While there is disagreement, most fusion
scientists will take the neutron problem and the D-T fuel,
because it is very difficult just to get D-T reactions to go.)
Another difficulty with the D-T reaction is that the tritium
is (weakly) radioactive, with a half-life of 12.3 years, so
that tritium does not occur naturally. Getting the tritium
for the D-T reaction is therefore another problem.
Fortunately you can kill two birds with one stone, and solve
both the neutron problem and the tritium-supply problem at
the same time, by using the neutron generated in the D-T
fusion in a reaction like n + Li6 -> He4 + T + 4.8 MeV.
This absorbs the neutron, and generates another tritium,
so that you can have basically a D-Li6 fuel cycle, with
the T and n as intermediates. Fusing D and T, and then
using the n to split the Li6, is easier than simply trying
to fuse the D and the Li6, but releases the same amount of
energy. And unlike tritium, there is a lot of lithium
available, particularly dissolved in ocean water.
Unfortunately you can't get every single neutron to stick
to a lithium nucleus, because some neutrons stick to other
things in your reactor. You can still generate as much
T as you use, by using "neutron multipliers" such as
Beryllium, or by getting reactions like
n + Li7 -> He4 + T + n (which propagates the neutron)
to occur. The neutrons that are lost are still a problem,
because they can induce radioactivity in materials that
absorb them. This topic is discussed more in Section 2.