Basic Math Problem of the Week 10/28/2017

[PF PotW Robot]

Here is this week's basic math problem. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods.

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 $2\sqrt{2n} \le k \le 3\sqrt{n}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[fresh_42]

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 $2\sqrt{2n} \le k \le 3\sqrt{n}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

Hint for the difficult part: Try to use the distance between the minimum and the zero of the parabola, $S_n := \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}$ for the estimation.

 

[QuantumQuest]

Just an idea for the right part of the inequality to be proved

Spoiler

Let $x_1$ and $x_2$ be the roots of the given quadratic equation with $x_1 = \dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}$
Using Vieta's formulas for quadratics ($ax^2 + bx + c = 0$) and applying to the given quadratic equation we have that $x_1 + x_2 = -\frac{b}{a} = k$ and $x_1\cdot x_2 = \frac{c}{a} = 2n$. So, $x_2 = \frac{2n}{\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}}$. Now, going to the inequality we're asked to prove we substitute $k$ for the sum of the two roots $x_1 + x_2$ and we have
$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}} + \frac{2n}{\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}} \leq 3\sqrt{n}$.
Now, to prove this right part of the asked inequality, we utilize that $\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}} < 2\sqrt{n}$. This can be easily proved using induction, so I omit the proof here. Returning to the right part of the asked inequality and doing the math we have

$\frac{(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}})^2 + 2n}{\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}} \leq \frac{6n}{2\sqrt{n}} = 3\sqrt{n}$ which is what is asked in the right part.

 

[Charles Link]

An excellent solution by @QuantumQuest ! I would like to make one minor correction though. In proving that $S_n+\frac{2n}{ S_n} \leq 3 \sqrt{n}$, I think in writing it as a fraction, you would need to have a lower limit on the denominator (rather than an upper limit) to prove the last "less than or equal" inequality. $\\$ Instead, let $S_n=2 \sqrt{n}-\delta_n$. Also let $A=\sqrt{n}$. Then we need to find the conditions on $\delta_n$ so that $2A-\delta_n+\frac{2A^2}{2A-\delta_n} \leq 3 A$. Multiplying through on both sides by $2A-\delta_n$, we get $(2A-\delta_n)^2+2A^2 \leq (3A)(2A-\delta_n)$. The result after a little algebra ( multiplying out, and subtracting, etc.) is $\delta_n \leq A$. Thereby $\delta_n \leq \sqrt{n}$ is the requirement for the inequality to hold. For the series $S_n$, when $n=1$, $\delta_1=1$ . For integers $n>1$, it looks as if it will clearly be the case that $\delta_n <\sqrt{n}$, but on this part (for $n>1$), I will forego a rigorous proof.