Prove 2 sqrt(x) > 3 - (1/x) for x > 1

[issacnewton]

Homework Statement

For $x> 1$, prove that $$2\sqrt{x} > 3 - \frac 1 x$$

Homework Equations

Definition of increasing function

The Attempt at a Solution

Let $f(x) = 2\sqrt{x} + \frac 1 x$ defined on domain $[1, \infty)$. Function is increasing if $f'(x) > 0$ on some interval $I$. This leads to $\frac{1}{\sqrt{x}} - \frac{1}{x^2} > 0$. So we have $\frac{1}{\sqrt{x}} > \frac{1}{x^2}$. Since $x \in [1, \infty)$, we can flip this $\sqrt{x} < x^2$. Therefore, $x < x^4$. Hence $x (x^3-1) > 0$. $\therefore x(x-1)(x^2+x+1) > 0$. Since $x > 0$ and $(x^2+x+1) > 0$ if $x \in [1, \infty)$, we have $x > 1$. So $f(x)$ is increasing on $(1,\infty)$. Its also known that $f(1) = 3$. How do I proceed from here on ?

Thanks

 

[mfb]

I would say you are done.

You showed equality for x=3 and that the difference between the two sides is strictly increasing for x>1.

You can use the mean value theorem to make the last step more formal.

 

[Ray Vickson]

Homework Statement

For $x> 1$, prove that $$2\sqrt{x} > 3 - \frac 1 x$$

Homework Equations

Definition of increasing function

The Attempt at a Solution

Let $f(x) = 2\sqrt{x} + \frac 1 x$ defined on domain $[1, \infty)$. Function is increasing if $f'(x) > 0$ on some interval $I$. This leads to $\frac{1}{\sqrt{x}} - \frac{1}{x^2} > 0$. So we have $\frac{1}{\sqrt{x}} > \frac{1}{x^2}$. Since $x \in [1, \infty)$, we can flip this $\sqrt{x} < x^2$. Therefore, $x < x^4$. Hence $x (x^3-1) > 0$. $\therefore x(x-1)(x^2+x+1) > 0$. Since $x > 0$ and $(x^2+x+1) > 0$ if $x \in [1, \infty)$, we have $x > 1$. So $f(x)$ is increasing on $(1,\infty)$. Its also known that $f(1) = 3$. How do I proceed from here on ?

Thanks

Your mode of reasoning is unsound. You want to know whether $f(x)$ is increasing on $(1,\infty)$. So you ultimately want to know if $\sqrt{x} < x^2$ on $(1,\infty)$. That is true if you can show that $\sqrt{1} \leq 1^2$ and $(\sqrt{x})' < (x^2)'$ on $(1,\infty)$.

Throughout, you have been essentially assuming what you want to prove and then showing at the end that this implies $x > 1$. In other words, if A = "$x > 1$" and B = "$f(x)$ is increasing", you have shown that $B \Rightarrow A$, but what what is needed is $A \Rightarrow B$. Logically there is a real difference.

 

[mfb]

Huh?

He calculated the derivative and showed that it is positive for x>1 in a direct way (although not with the best possible phrasing). What is unsound about that?

 

[issacnewton]

Hello Ray, may be the initial statement of the problem is confusing. But I think I proved that $f(x)$ is increasing on $(1,\infty)$ without any assumptions.
mfb, I don't see how intermediate value theorem would be useful here. To reach eventual conclusion, I have to prove that $\forall~x \in (1,\infty), f(x) > f(1)$. I tried to assume the negation, $f(x) \leqslant f(1)$. I reached the contradiction for $f(x) = f(1)$. I am trying how to get contradiction for $f(x) < f(1)$. Would this be correct approach ?

 

[mfb]

mfb, I don't see how intermediate value theorem would be useful here.

It tells you that, for every x₀>1 there is a value 1<x<x₀ where f'(x)*...=..., and you can use this to show that f(x₀)>3.

 

[issacnewton]

You have used derivative in the statement. Intermediate value theorem does not talk about any derivatives.

 

[Ray Vickson]

Hello Ray, may be the initial statement of the problem is confusing. But I think I proved that $f(x)$ is increasing on $(1,\infty)$ without any assumptions.
mfb, I don't see how intermediate value theorem would be useful here. To reach eventual conclusion, I have to prove that $\forall~x \in (1,\infty), f(x) > f(1)$. I tried to assume the negation, $f(x) \leqslant f(1)$. I reached the contradiction for $f(x) = f(1)$. I am trying how to get contradiction for $f(x) < f(1)$. Would this be correct approach ?

No, you showed that if $f(x)$ is increasing then $x > 1$. In your case the steps are almost reversible, so doing what you did is harmless, but fundamentally it is a mode of demonstration that you should try to avoid. In your case proving "A implies B" and "B implies A" are essentially reversed arguments of each other, and that is why you can get away with it here. However, logically speaking, "A implies B" and "B implies A" are very different and are sometimes not the same at all: you can have one without the other in some cases.

Sometimes when we solve a problem we essentially do what you did (at least in the exploratory phase); that is, we sometimes start with what we are trying to prove and then work backwards towards the "hypothesis". If we do that it is to gain insight, etc. However, when we write up the final solution we should throw away all those sheets of paper and re-write the argument in the proper fashion. Some instructors would accept what you did, while others would mark it incorrect. Better to be safe.

 

[issacnewton]

Ray, ok, I will be careful about my arguments.

 

[mfb]

You have used derivative in the statement. Intermediate value theorem does not talk about any derivatives.

I said mean value theorem, not intermediate value theorem.

 

[issacnewton]

Ok, I need to prove that $\forall~x \in (1, \infty), f(x) > f(1)$. We let $x \in (1, \infty)$ be arbitrary. Now consider the interval $[1, x]$. $f(x)$ is continuous on $[1,x]$ and differentiable on $(1,x)$. So there is $c \in (1,x)$ such that $f'(c) = \frac{f(x)-f(1)}{x-1}$. Since $c >1$, and $f(x)$ is increasing on $(1,\infty)$, we have $f'(c) > 0$. Since $x > 1$, we have $f(x) >f(1)$. Hence $2\sqrt{x} + \frac 1 x > 3$, or equivalently, $2\sqrt{x} > 3 - \frac 1 x$. Since $x$ is arbitrary, this is true for all $x > 1$. I hope the arguments are correct now.

 

[mfb]

Correct.