Let f(x) = limn→∞[∏nk=1(x+nk)n!(∏nk=1(x2+n2k2))]1/n, for all x>0. Then f(1/2)≥f(1), f(1/3)≤f(2/3), f'(3)f(3)≥f'(2)f(2), f'(2)≤0

f(x)=limn→∞[∏nk=1(x+nk)n!(∏nk=1(x2+n2k2))]1/n
f(x)=limn→∞[∏nk=1(xn+k)n!(∏nk=1(x2n2+k2))]1/n=limn→∞xn[∏nk=1(1+xn/k)∏nk=1(1+k2x2n2)]−1/n
f(x)=limn→∞xn[∑nk=1log(1+kx/n)−∑nk=1log(1+k2x2n2)]
f(x)=limn→∞xn[∑nk=1(kx/n)−∑nk=1(k2x2n2)/2]=x∫10(1+xy)dy−x∫10(1+x2y2)dy
Let xy=t
b(f(x))=∫x0(1+t)dt−∫x0(1+t2)dt
f'(x)f(x)=(1+x1+x2)
f'(2)f(2)=(35)<0⇒f'(2)<0
f'(3)f(3)=(410)=(25)⇒f'(2)f(2)≥f'(3)f(3)
Now f'(x)f(x)>0 in (0, 1) and f'(x)f(x)<0 in (1,∞)
f(x) is increasing in (0, 1) decreasing in [1,∞) (as f(x) is positive)
Hence f(1)≥f(1/2) and f(1/3)≤f(2/3)