Selection and Training Session

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

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



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

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

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