MHB How Do You Prove the Root Constraints of a Quadratic Equation Given a Summation?

[anemone]

Here is this week's POTW:

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One of the roots of the quadratic equation $x^2-kx+2n=0$ is equals to $\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}$, where $n$ is a positive integer.

Prove that $2\sqrt{2n} \le k \le 3\sqrt{n}$.

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[anemone]

Congratulations to the lfdahl for his correct solution(Smile), and you can read the suggested solution as follows:

Spoiler

Let $$u=\sum_{a=1}^{n}\frac{1}{\sqrt{a}}$$ and $$v=\frac{u}{\sqrt{n}}$$ so we have $$k=u+\frac{2n}{u}$$.

Note that for all $$a\in [1,\,n]$$, it must be true that $$\frac{1}{\sqrt{a}}\ge \frac{1}{\sqrt{n}}$$. From here we get $$u\ge \frac{n}{\sqrt{n}}$$ and that simply is $$u \ge \sqrt{n}$$.

Now, for all $$a\in (1,\,n]$$, $$\frac{1}{\sqrt{a}}\le \int_{a-1}^{a} \,\frac{dx}{\sqrt{x}}$$ and so we get

$$\sqrt{n}\le u \le 1+\int_{1}^{n} \,\frac{dx}{\sqrt{x}}$$

$$\sqrt{n}\le u \le 2\sqrt{n}-1 \le 2\sqrt{n}$$

$$1\le \frac{u}{\sqrt{n}} \le 2$$

$$1\le v \le 2$$

So we get $v-2 \le 0$ and $v-1 \ge 0$ and thus

$(v-2)(v-1) \le 0$

$v^2-3v+2 \le 0$

$v+\dfrac{2}{v} \le 3$

Since we have $x+\dfrac{2}{x}\ge 2\sqrt{2}$ for all $x>0$, we get

$2\sqrt{2} \le v+\dfrac{2}{v} \le 3$

$2\sqrt{2n}\le u+\dfrac{2n}{u}\le 3\sqrt{n}$

$2\sqrt{2n} \le k \le 3\sqrt{n}$ (Q.E.D.)