Finding if sentence is true using induction: $\sqrt1 + \sqrt2 +\dots+\sqrt{n} \le \frac23(n+1)\sqrt{n+1}$

Question

I have to prove

$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le \frac{2}{3}*(n+1)\sqrt{n+1}$

by using math induction.

First step is to prove that it works for n = 1 , which is true. Next step is to prove it for n + 1. We can rewrite the formula using

$\sum_{i=1}^{n+1} \sqrt{i}= \sum_{i=1}^{n}\sqrt{i} + \sqrt{i+1}$

and we can substitute sum

$\frac{2}{3}(n+1)\sqrt{n+1} +\sqrt{n+1} \le \frac{2}{3}(n+2)\sqrt{n+2}$

we can transform the left side into

$\sqrt{n+1}(\frac{2}{3}(n+1)+1)$

but how to I further transform the formula in order to find if the sentence is true?

Thanks for all help!

Answer 1

You started off well and almost got it. Here's a hint, where LHS and RHS are respectively the left and right hand sides of the formula we want:

$$\begin{align}\mathrm{LHS}^2&=\left(\frac23\right)^2(n+1)\left(n+\frac52\right)^2\\ &=\left(\frac23\right)^2(n^3+6 n^2+11.25n+6.25) \\ &<\left(\frac23\right)^2(n^3+6n^2+12n+8)&&=\left(\frac23\right)^2(n+2)^3\\ &&&=\mathrm{RHS}^2\\ \mathrm{LHS}&<\mathrm{RHS} \end{align}$$

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An alternative approach, with some calculus.

$$\begin{align} \ln(\mathrm{LHS})&=\ln\frac23+\frac12\ln(n+1)+\ln(n+\frac52)\\ \ln(\mathrm{RHS})&=\ln\frac23+\frac32\ln(n+2)\\ \\ f(n)&=\ln(\mathrm{RHS})-\ln(\mathrm{LHS})\\ &=\frac32\ln(n+2)-\frac12\ln(n+1)-\ln(n+\frac52)\\ f'(n)&=\frac{\frac32}{n+2}-\frac{\frac12}{n+1}-\frac1{n+\frac52}\\ &=\frac{-\frac34}{(n+2)(n+1)(n+\frac52)}\\ \end{align}$$

So $f'(n)$ is never $0$, which means that $f$ has no turning points but we can show that $f(1)=0.049\ldots$ and hence $\ln(\mathrm{RHS})-\ln(\mathrm{LHS})>0$

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Let me know if you've spotted an error.

-Jam

Answer 2

To finish your proof by induction we could use the fact the following inequalities are equivalent to each other \begin{align*} \frac23(n+1)^{3/2}+\sqrt{n+1} &\le \frac23(n+2)^{3/2}\\ \frac32\sqrt{n+1} &\le (n+2)^{3/2} - (n+1)^{3/2}\\ \frac32\sqrt{n+1} &\le (\sqrt{n+2} - \sqrt{n+1}) (n+2 + \sqrt{(n+2)(n+1)} + n+1)\\ \frac32\sqrt{n+1}(\sqrt{n+2} + \sqrt{n+1}) &\le n+2 + \sqrt{(n+2)(n+1)} + n+1\\ \frac32(n+1 + \sqrt{(n+2)(n+1)}) &\le 2n+3 + \sqrt{(n+2)(n+1)}\\ \frac12\sqrt{(n+2)(n+1)} &\le \frac{n+3}2 \end{align*} I have:
Used $a^3-b^3=(a-b)(a^2+ab+b^2)$ for $a=\sqrt{n+2}$ $b=\sqrt{n+1}$
Multiplied both sides by $\sqrt{n+2}+\sqrt{n+1}$ to use that $(\sqrt{n+2} - \sqrt{n+1})(\sqrt{n+2} + \sqrt{n+1}) = (n+2)-(n+1)=1$.

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If you already know how to integrate, you can simply use that $$\newcommand{\dx}{\; \mathrm{d}x}\sqrt k \le \int_{k}^{k+1} \sqrt x \dx.$$ Simply by adding these inequalities together for $k=0$ to $n$ you get $$ \sum_{k=0}^n \sqrt n \le \int_0^{n+1} \sqrt x \dx = \frac{(n+1)^{3/2}}{3/2} = \frac23 (n+1)\sqrt{n+1}.$$

This can be also visualized by noticing that one side is are of rectangle under the curve $y=\sqrt x$ and the other is the whole area under the curve.

I will add the following - somewhat similar - picture which might help (it was taken from this post):