TITLE: Prove Inequality without induction.
QUESTION [2 upvotes]: I showed this inequality by induction. I want other methods to prove it.
$(\frac{2n}{3}+\frac{1}{3})\sqrt{n}\leq \sum_{k=1}^{n}\sqrt{k}\leq (\frac{2n}{3}+\frac{1}{2})\sqrt{n}$
Thank

REPLY [1 votes]: ($\frac{2n}{3}$+$\frac{1}{3})$$\sqrt{n}$$\leq $$\sum_{k=1}^{n}$$\sqrt{k}$$\leq $$(\frac{2n}{3}$+$\frac{1}{2})$$\sqrt{n}$ ?  
By these inequalities    :
$\forall$ k $\geq$ 1 : $\frac{2}{3}$[$k^\frac{3}{2}$-$(k-1)^\frac{3}{2}$]+$\frac{1}{3}$[$k^\frac{1}{2}$-$(k-1)^\frac{1}{2}$]$\leq$ $k^{\frac{1}{2}}$$\leq$ $\frac{2}{3}$[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+$\frac{1}{2}$[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$]   $(*)$
and we sum by telescoping we get the result. 
show (*) 
3$a_{k}$=2[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]+[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$] - 3$k^{\frac{1}{2}}$= 2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}(2(k-1)+1)$ $\\$
                                                              3$a_{k}$=2$k^{\frac{1}{2}}$(k-1)-$(k-1)^{\frac{1}{2}}$(2k-1)$\leq $0 because 4k$(k-1)^{2}$$\leq$ (k-1)$(2k-1)^{2}$ indeed 4k(k-1)-(2k-1)²= -1$\leq $0
6$b_{k}$=6$k^{\frac{1}{2}}$-4[$k^{\frac{3}{2}}$-$(k-1)^{\frac{3}{2}}$]-3[$k^{\frac{1}{2}}$-$(k-1)^{\frac{1}{2}}$]
6$b_{k}$=$(k-1)^{\frac{1}{2}}$(4k-1)-$k^{\frac{1}{2}}$(-3+4k)$\leq$ 0 because (k-1)$(4k-1)^{2}$$\leq$ k$(4k-3)^{2}$
indeed (k-1)$(4k-1)^{2}$ - k$(4k-3)^{2}$= -1$\leq$0
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