With the notation of exerc. 6 show that
μ1G1+μ2G2+⋯+μnGn≤λ1a1+λ2a2+⋯+λnan
provided that λk≥0 for each k and
μk=k((λkβk)1/k−βk+11/k)for 1≤k≤n
with βk≥0 for all k, β1≤1 and βn+1=0 (method of exerc. 8).
In particular, for suitable choices of λk and βk one has
G1+G2+⋯+Gn+nGn≤2a1+(23)2a2+⋯+(nn+1)nan
and in particular (Carleman’s inequality)
G1+G2+⋯+Gn<e(a1+a2+⋯+an);
eAn≥ΓneGn/Γn
where Γn=(G1+G2+⋯+Gn)/n (strengthening of the Carleman inequality);
GnAn≥21(GnΓn+ΓnGn)
if the sequence (ak) is increasing:
nGn−mGm≤am+1+am+2+⋯+an.