Răspuns:
Banuiesc ca ai rezolvat cazul n=1
Luam k pentru care este adevarata relatia
[tex]\frac{1}{2}\times\frac{3}{4}...\frac{2k-1}{2k}\leq \frac{1}{\sqrt{3k+1}} \\ \\ \frac{1}{2}\times\frac{3}{4}...\frac{2k-1}{2k}\times\frac{2k+1}{2k+2}\leq \frac{1}{\sqrt{3k+1}}\times\frac{2k+1}{2k+2}[/tex]
Trebuie sa aratam:
[tex]\frac{1}{\sqrt{3k+1}}\times\frac{2k+1}{2k+2}\leq\frac{1}{\sqrt{3k+4}} \\ \\ (2k+1)\sqrt{3k+4}\leq (2k+2)\sqrt{3k+1} \\ \\ (4k^2+4k+1)(3k+4)\leq (4k^2+8k+4)(3k+1) \\ \\ 12k^3+12k^2+3k+16k^2+16k+4\leq 12k^3+24k^2+12k+4k^2+8k+4 \\ \\ 12k^3+28k^2+19k+4\leq 12k^3+28k^2+20k+4 \\ \\ k\geq 0[/tex]
Ceea ce este adevarat
