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

• issacnewton
In summary: 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.In this case, you did not prove that ##f(x)## is increasing on ##(1,\infty)##. You only proved that if ##f(x)## is increasing then ##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

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.

IssacNewton said:

## 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.

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?

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 ?

IssacNewton said:
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.

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

IssacNewton said:
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.

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

IssacNewton said:
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.

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.

Correct.

## What does "Prove 2 sqrt(x) > 3 - (1/x) for x > 1" mean?

The statement is an inequality that needs to be proven to be true for all values of x greater than 1. In other words, we need to show that the square root of 2x is always greater than 3 minus 1 over x when x is greater than 1.

## Why is it important to prove this inequality?

Proving this inequality is important because it helps establish a mathematical relationship between two expressions. It also allows us to make conclusions about the behavior of the expressions for different values of x, which can be useful in solving other problems.

## What are the steps involved in proving this inequality?

The steps involved in proving this inequality may vary, but they typically involve simplifying the expressions, setting them equal to each other, and then using algebraic manipulations to show that one side is always greater than the other for x > 1.

## What are some common strategies for proving inequalities?

Some common strategies for proving inequalities include using algebraic manipulations, applying mathematical properties such as the distributive property, using the properties of inequalities (such as multiplying or dividing by a positive number), and using substitution to simplify the expressions.

## Can this inequality be proven using other methods besides algebraic manipulation?

Yes, this inequality can also be proven using graphical methods, such as graphing both expressions on a coordinate plane and showing that the graph of 2 sqrt(x) is always above the graph of 3 - (1/x) for x > 1. Additionally, it can also be proven using calculus methods, such as taking the derivatives of both expressions and comparing them.

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