<s>[INST] 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 $1$"?
 [/INST] 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.
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