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

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 $$1$$"?

• On my calculator the sum is $1.019877344$. Commented Jan 19, 2020 at 0:18
• @OscarLanzi Yes, I know what the sum is. I'm looking for a way to recognize that it is larger than 1 without actually adding all of the numbers together. Commented Jan 19, 2020 at 0:19

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.

• This is pretty much exactly what I was looking for. Thanks! Commented Jan 19, 2020 at 0:16
• You could mention that this is the inequality between harmonic and arithmetic mean: $\frac{2}{1/a+1/b} < \frac{a+b}{2}$. Commented Jan 20, 2020 at 7:55
• @mweiss if this solution is what you are looking for, then I think it should be chosen as an answer. Commented Jan 29, 2020 at 3:18

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

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

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

• This is also great. Thank you! Commented Jan 19, 2020 at 1:17

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

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

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

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

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

• My analysis uses C-S with pairs of terms. +1 for getting a bound actually greater than $1$. Commented Jan 19, 2020 at 12:10