“…2m) 3/2 (n 2 − 2m) 3/2 + 4mn − n3 and, if 0 m n 2 /4, then(2m) 3/2 (n 2 − 2m) 3/2 + 4mn − n 3 .In other words,F (n, m) = max{(2m) 3/2 , (n 2 − 2m) 3/2 + 4mn − n 3 }.Lemma 4 implies that, for all n and m,C(n, m) m √ 8m + 1 − m (2m) 3/2 ;Lemma 5 implies that, for m n 2 /4, S(n, m) (n 2 − 2m) 3/2 + 4mn − n 3 , and so, in view of (4), the second inequality in(8)is proved.Proof of the first inequality in(8): To prove the first inequality in(8), observe that, by Proposition 3, we have (2m) 3/2 − 3m C(n, m), and so, if m n 2 /4, then F (n, m) − 4m C(n, m).…”