[gon]

ขอสักข้อล่ะกันครับ.

ข้อ 6 จะแสดงว่า
\((1+2+3+...+n)(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}) > (n+1)^2 , ทุก\, n \geq 6\)

เนื่องจาก \(L.H.S. = (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})(\frac{n^2+n}{2}) + (\frac{1}{5}+\frac{1}{6}+...+\frac{1}{n})(\frac{n^2+n}{2})\)
\( = \frac{25}{12}(\frac{n^2+n}{2}) + (\frac{1}{5}+\frac{1}{6}+...+\frac{1}{n})(\frac{n^2+n}{2})\)
\( > 2(\frac{n^2+n}{2}) + (\frac{1}{5}+\frac{1}{6}+...+\frac{1}{n})(\frac{n^2+n}{2})\)
\( > n^2 + n + (\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n})(\frac{n^2+n}{2}) จำนวน \, n - 4 พจน์\)
\( = n^2 + n + \frac{(n-4)(n+1)}{2}\)
\( = \frac{3n^2-n-4}{2} \geq n^2 + 2n + 1\)
\( \Leftrightarrow 3n^2 - n - 4 \geq 2n^2 + 4n + 2\)
\( \Leftrightarrow n^2 - 5n - 6 \geq 0\)
\( \Leftrightarrow (n-6)(n+1) \geq 0\)
\( \Leftrightarrow n \geq 6\)