Đặt vế trái của Bất đẳng thức la A
\(A< \frac{1}{8}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{40}+\frac{1}{40}+\frac{1}{40}.\)
\(A< \frac{1}{8}+\frac{3}{10}+\frac{3}{40}=\frac{3}{10}< \frac{5}{10}=\frac{1}{2}\)
Ta thấy: \(\frac{1}{8}< \frac{1}{2}\)
\(\frac{1}{11}< \frac{1}{2}\)
\(\frac{1}{12}< \frac{1}{2}\)
\(\frac{1}{13}< \frac{1}{2}\)
\(\frac{1}{41}< \frac{1}{2}\)
\(\frac{1}{42}< \frac{1}{2}\)
\(\frac{1}{43}< \frac{1}{2}\)
=> \(\frac{1}{8}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{41}+\frac{1}{42}+\frac{1}{43}< \frac{1}{2}\)