Answer: b) yes.

We use the following well-known

Lemma. Let $a, b, c, d$ be the lengths of the sides of some quadrilateral, and $S$ be its area. Then $2S \le ab + cd$ and $2S \le ac + bd$.

Let now $a, b, c, d, e$ be the lengths of the sides of the given pentagon, $f$ be the length of one of its diagonals (see the Fig.), $S$ be its area. Then $2S \le ab + cf + de < ab + c(d + e) + de$, i.e. $2S \le (ab + cd + de) + ce$. Note that all summands in parentheses are the products of the neighboring sides of the pentagon, while $ce$ is the product of non-neighboring sides.

Summing all similar inequalities which we can obtain by the cyclic permutation $a \to b \to c \to d \to e \to a$, we get

$$10S \le 3B + C, \quad (1)$$

where $B$ is the sum of all five pairwise products of the neighboring sides, $C$ is the sum of all five pairwise products of the non-neighboring sides.

Further, $2S \le ac + bf + de < ac + b(d+e) + de = (ac + bd + be) + de$. In the same way we obtain from the last inequality the following one

$$10S \le 3C + B. \quad (2)$$

Summing (1) and (2) gives the required inequality.