, 2008

Solution:
If $1 \geq a + b + c$ then we write the given inequality equivalently as
$$\begin{gathered} 1 - (a + b + c) + d[(a + b + c) - 1] + a b + b c + c a > 0 \\ \Leftrightarrow [1 - (a + b + c)](1 - d) + a b + b c + c a > 0 \end{gathered}$$
which is of course true.

If instead $a + b + c > 1$, then $d(a + b + c) > d$ i.e.
$$d a + d b + d c > d$$
We are going to prove that also
$$1 + a b + b c + c a > a + b + c$$
thus adding (1) and (2) together we'll get the desired result in this case too.

For the truth of (2):
If $1 \geq a + b$, then we rewrite (2) equivalently as
$$\begin{gathered} 1 - (a + b) + c[(a + b) - 1] + a b > 0 \\ \quad \Leftrightarrow [1 - (a + b)](1 - c) + a b > 0 \end{gathered}$$
which is of course true.

If instead $a + b > 1$, then $c(a + b) > c$, i.e.
$$c a + c b > c$$
But it is also true that
$$1 + a b > a + b$$
because this is equivalent to $(1 - a) + b(a - 1) > 0$, i.e. to $(1 - a)(1 - b) > 0$ which holds. Adding (3) and (4) together we get the truth of (2) in this case too and we are done.

You can instead consider the following generalization: