Is there an easy way to see that ${1\over5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} > 1$?

Question

The sum $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}$$ is just a bit larger than $1$. Is there some clever way to show this other than to add the fractions together by brute-force? For example, is there some way to group terms together and say something like "These terms sum to more than $\frac{1}{3}$, these terms sum to more than $\frac{1}{2}$, and these terms sum to larger than $\frac{1}{6}$, so the whole thing sums to more than $1$"?

Answer 1

For positive, unequal $a$ and $b$:

$\dfrac1a+\dfrac1b=\dfrac{a+b}{ab}>\dfrac4{a+b}$

because $(a+b)^2>4ab$ (the difference between these is $(a-b)^2$). So,

$\dfrac15+\dfrac17>\dfrac4{12}=\dfrac13$

$\dfrac19+\dfrac1{11}>\dfrac4{20}=\dfrac1{5}$

$\dfrac18+\dfrac1{12}>\dfrac4{20}=\dfrac1{5}$

When these inequalities are put into the given sum the claimed bound follows.

Answer 2

Since $y=\frac{1}{x}$ is convex we have:-

$\dfrac15+\dfrac16+\dfrac17>\dfrac36=\dfrac12$

$\dfrac18+\dfrac19+\dfrac1{10}+\dfrac1{11}+\dfrac1{12}>\dfrac5{10}=\dfrac12$

Answer 3

$\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}=\dfrac{6}{20}$

$\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{3}{12}=\dfrac{5}{20}$

$\dfrac{1}{7}+\dfrac{1}{8}>\dfrac{2}{8}=\dfrac{5}{20}$

$\dfrac{1}{9}+\dfrac{1}{11}=\dfrac{20}{99}>\dfrac{4}{20}$

Answer 4

By C-S we obtain: $$\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12}=$$ $$=\frac{17}{60}+\frac{17}{66}+\frac{17}{70}+\frac{17}{72}\geq\frac{17\cdot4^2}{60+66+70+72}=\frac{68}{67}>1.$$