<s>[INST] Prove that $\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$ How do I prove $$\frac 12\leq \frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n}$$ 
without using induction? Note that clearly $n\neq 0$
Thanks for any help!!
 [/INST] The reverse inequality is true for $n≥1$.
If $1≤k≤n$ then $\frac{1}{n+k}≥\frac{1}{n+n}$. Therfore, you get :

$$\frac{1}{n+1}+\frac {1}{n+2}+\frac{1}{n+3}...+\frac{1}{n+n} ≥ \frac{1}{n+n}+\frac {1}{n+n}+\frac{1}{n+n}...+\frac{1}{n+n} = n \cdot \frac{1}{n+n} = \frac{1}{2}$$

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