Prove that $\sqrt1+\sqrt2+\sqrt3+....+\sqrt n<n\sqrt{\frac{n(n+1)}{2}}$

Question

$Q.$ Prove that $$\underbrace{\sqrt1+\sqrt2+\sqrt3+....+\sqrt n}_{\alpha}<n\sqrt{\underbrace{\frac{n(n+1)}{2}}_\beta}$$

Basically we need to prove that $$\alpha^2<n^2\beta$$

And $$\alpha^2=\beta+2(\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+......+\sqrt{(n-1)n})$$ using AM-GM on $(n-1)$ and $n$ $$\frac{2n-1}{2}>\sqrt{(n-1)(n)}$$

So , $$\alpha^2<\beta+\sum_{i=2}^{n}(2n-1)$$ $$\alpha^2<\beta-1+\sum_{i=1}^{n}2n-1$$ $$\alpha^2<\beta-1-n+n(n+1)$$ $$\alpha^2<\beta+(n+1)(n-1)$$

Now I'm not able to solve further . Can you give some hints ?

Answer 1

Since you asked for a solution using AM-GM:

\begin{multline*} \sqrt k + \sqrt{n + 1 - k} = \sqrt{(\sqrt k + \sqrt{n + 1 - k})^2} = \sqrt{n + 1 + 2\sqrt{k(n + 1 - k)}} \\ ≤ \sqrt{n + 1 + k + (n + 1 - k)} = 2\sqrt{\frac{n + 1}{2}}. \end{multline*}

The (strengthened) result follows by summing from $k = 1$ to $n$.

Answer 2

By C-S $$\sum_{k=1}^n\sqrt{k}\leq\sqrt{\sum_{k=1}^n1^2\sum_{k=1}^nk}=\sqrt{n\cdot\frac{n(n+1)}{2}}\leq n\sqrt{\frac{n(n+1)}{2}}$$ The equality does not occur in the general

Answer 3

Note that the (intermediate) Cauchy–Schwarz bound given by Michael Rozenberg is much better than what you wanted: $$ \sum_{k=1}^n \sqrt{k} \leq n\sqrt{\frac{n+1}{2}} \tag{1} $$

as it grows as $n^{3/2}$ instead of $n^2$.

You can prove a similar bound as follows. $$\begin{align} \sum_{k=1}^n \sqrt{k} &= \sum_{k=1}^n \int_{k}^{k+1} \sqrt{k}\, dx \leq \sum_{k=1}^n \int_{k}^{k+1}\sqrt{x}\, dx = \int_{1}^{n+1}\sqrt{x}\, dx = \frac{2}{3}((n+1)^{\frac{3}{2}}-1) \\ &\leq \frac{2}{3}(n+1)^{\frac{3}{2}} \tag{2} \end{align}$$ This is a slightly better bound, as $\frac{2}{3} < \frac{1}{\sqrt{2}}$. In particular, it is better than the bound $n\sqrt{\frac{n+1}{2}}$ for all $n \geq 17$.

Answer 4

$\sqrt1+\sqrt2+\sqrt3+....+\sqrt n<n\sqrt{\frac{n(n+1)}{2}}$

Can't resist alternative approach, which uses induction.

Assertion true for $n = 2$.

Assume assertion true for $N$.

As $N$ goes to $N+1$, the LHS increases by
$\sqrt{N+1}$
while the RHS increases by
$\displaystyle \left[(N+1)\sqrt{\frac{(N+1)(N+2)}{2}} - N\sqrt{\frac{N(N+1)}{2}}\right].$

Therefore, the question reduces to showing that for $~N \in \Bbb{Z_{\geq 2}}~~$:

$$\sqrt{N+1} < \left[(N+1)\sqrt{\frac{(N+1)(N+2)}{2}} - N\sqrt{\frac{N(N+1)}{2}}\right].\tag1 $$

In (1) above, the common factor of $\displaystyle \sqrt{N+1}$ may be removed from both sides.

Therefore, the problem reduces to showing that for $~N \in \Bbb{Z_{\geq 2}}~~$ :

$$ 1 < \left[(N+1)\sqrt{\frac{(N+2)}{2}} - N\sqrt{\frac{N}{2}}\right].\tag2 $$

The RHS of (2) above may be re-expressed as

$$\sqrt{\frac{(N+2)}{2}} + N\left[\sqrt{\frac{(N+2)}{2}} - \sqrt{\frac{N}{2}}\right]. \tag3 $$

Symbolizing the expression in (3) above as $\displaystyle A + N(B)$, clearly $A > 1$ and $B > 0$. Therefore, the inequality asserted in (2) above is established.

This concludes the induction step. Therefore, the original assertion is true, by induction.

Answer 5

Here's what I did , $$\alpha^2=(\sqrt{1}+\sqrt{2}+\sqrt{3}+......+\sqrt{n})^2\Rightarrow$$ $$\color{red}{\sqrt1.\sqrt1}+\sqrt1.\sqrt2+\sqrt1.\sqrt3+.....+\sqrt1.\sqrt n\\\sqrt2.\sqrt1+\color{red}{\sqrt2.\sqrt2}+\sqrt2.\sqrt3+.....+\sqrt2.\sqrt n\\\sqrt3.\sqrt1+\sqrt3.\sqrt2+\color{red}{\sqrt3.\sqrt3}+.....+\sqrt3.\sqrt n\\.\\.\\.\\\sqrt n.\sqrt1+\sqrt n.\sqrt2+\sqrt n.\sqrt3+.....+\color{red}{\sqrt n.\sqrt n} $$

Now apply AM-GM on each term , $$\alpha^2<\color{red}{(1+2+3+......+n)}+2\left(\frac{1+2}{2}+\frac{1+3}{2}+\frac{1+4}{2}+.......\right)$$OR $$\alpha^2<(1+2+3+.....+n)+(n-1)(1+2+3+......+n)$$ $$\alpha^2<n(1+2+3+.....+n)$$

since , $$n^2(1+2+3+.....+n)>n(1+2+3+......+n)$$ then $$(\sqrt1+\sqrt2+\sqrt3+......+\sqrt n)^2<n^2(1+2+3+......+n)$$

Hence $$\sqrt1+\sqrt2+\sqrt3+.......+\sqrt n<n\sqrt{\frac{n(n+1)}{2}}$$