Prove by induction that $\sqrt{1}+\sqrt{2}+...+\sqrt{n} \geq \frac{2}{3}n\sqrt{n}$

Question

The base step is pretty obvious: $1 \geq \frac{2}{3}$.

Then we assume that $P(k)$ is true for some $k \in \mathbb{Z}^{+}$ and try to prove $P(k+1)$. So I have

$ \sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}k\sqrt{k}+\sqrt{k+1}$

by the induction hypothesis. But I'm not too sure how to proceed to prove that this is also greater than $\frac{2}{3}(k+1)\sqrt{k+1}$.

Would appreciate any help!

Answer 1

You hope to have $$\frac23k\sqrt k+\sqrt{k+1}\stackrel?\ge \frac23(k+1)\sqrt{k+1} $$ or equivalently after simple transformations, $$\frac23k\sqrt k+\sqrt{k+1}\stackrel?\ge \frac23k\sqrt{k+1} +\frac23\sqrt{k+1},$$ $$\frac13\sqrt{k+1}\stackrel?\ge \frac23k\sqrt{k+1}-\frac23k\sqrt k,$$ $$\sqrt{k+1}\stackrel?\ge 2k(\sqrt{k+1}-\sqrt k).$$ A good trick when seeing differences of square roots is often to multiply with their sum, so here $$\sqrt{k+1}(\sqrt{k+1}+\sqrt k)\stackrel?\ge 2k(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)=2k((k+1)-k)=2k.$$ And now the claim is clear as we indeed have $$2k=\sqrt k(\sqrt k+\sqrt k)<\sqrt{k+1}(\sqrt{k+1}+\sqrt k). $$

Answer 2

For $P(k+1)$:
$\sqrt{1}+\sqrt{2}+...+\sqrt{k} + \sqrt{k+1} \geq \frac{2}{3}(k+1)\sqrt{k+1}$
$\frac 23k\sqrt k+\sqrt {k+1}\geq \frac 23k\sqrt k+\frac 23\sqrt {k+1}$
$\frac 13\sqrt{k+1}\geq 0$

Does this help?

Answer 3

If you can drop "by induction" there is another way to show the inequality.

At least, it shows how "others" may invent such inequalities:

$$\frac{2}{3}n\sqrt{n} \leq \sum_{i=1}^n \sqrt{i} \Longleftrightarrow \color{blue}{\frac{2}{3} \leq} \frac{1}{n\sqrt{n}}\sum_{i=1}^n \sqrt{i} = \color{blue}{\sum_{i=1}^n \sqrt{\frac{i}{n}}\cdot \frac{1}{n}}$$

The $\color{blue}{\mbox{blue}}$ sum is a Riemann sum for $\int_0^1 \sqrt{x}\; dx$ which can be estimated using the fact that $\sqrt{x}$ is strictly increasing:

• $\int_{\frac{i-1}{n}}^{\frac{i}{n}} \sqrt{x}\; dx < \int_{\frac{i-1}{n}}^{\frac{i}{n}} \sqrt{\frac{i}{n}}\; dx = \sqrt{\frac{i}{n}}\cdot \frac{1}{n}$ for $i=1, \ldots , n$

$$\color{blue}{\sum_{i=1}^n \sqrt{\frac{i}{n}}\cdot \frac{1}{n} >} \sum_{i=1}^n \int_{\frac{i-1}{n}}^{\frac{i}{n}} \sqrt{x}\; dx = \int_0^1 \sqrt{x}\; dx = \color{blue}{\frac{2}{3}}$$

Answer 4

Need to show:

$(2/3)k√k+\sqrt{k+1} \ge$

$ (2/3)(k+1)\sqrt{k+1}$, or

$\sqrt{k+1} \ge (2/3)[(k+1)^{3/2}-k^{3/2}]$.

$\displaystyle {\int_{k}^{k+1}} x^{1/2}dx \lt (k+1)^{1/2}(1)$

$\displaystyle {\int_{k}^{k+1}}x^{1/2}dx =$

$ (2/3)[(k+1)^{3/2}-k^{3/2}]$.

Hence

$(k+1)^{1/2} \gt \displaystyle {\int_{k}^{k+1}} x^{1/2}dx=$

$(2/3)[(k+1)^{3/2}-k^{3/2}].$

Answer 5

For the induction step, you want to show that: $$ \frac{2k\sqrt{k} + 3\sqrt{k+1}}{3} \geq \frac{2(k+1)\sqrt{k+1}}{3} \\ 2k\sqrt{k} + 3\sqrt{k+1} \geq 2k\sqrt{k+1} + 2\sqrt{k+1}\\ $$ Working backwards: $$ 2k\sqrt{k} + \sqrt{k+1} \geq 2k\sqrt{k+1} \\ 4k^2 \times k \geq (4k^2 - 4k+1)(k+1) = 4k^3 - 4k^2 + k + 4k^2 - 4k+ 1 = 4k^3 - 3k + 1 $$ The rest should follow since $k \geq \frac1{3}$, since the induction is over the positive integers.