Is Indirect Proof Easier for Proving Inequality Involving Real Numbers?

[tehdiddulator]

Homework Statement

Let x $\in$ ℝ
Prove that if 3x$^{4}$+1≤x$^{7}$+x$^{3}$, then x > 0

Homework Equations

None

The Attempt at a Solution

Assume
3x$^{4}$+1≤x$^{7}$+x$^{3}$
then 0 ≤ -3x$^{4}$-1≤x$^{7}$+x$^{3}$
Then I assumed that each was greater than or equal to 0, which I thought gave the desired result. No examples in the book to really guide me...any help would be greatly appreciated.

 

[clamtrox]

0 ≤ -3x$^{4}$-1

That can't be right, no matter what value x has.



You might want to think it like this instead: If x≤0, then x⁷+x³ is always smaller than 1+3x⁴. Why is this?

 

[tehdiddulator]

Would it be because both powers of x are odd, which will mean it is always negative? So proving the contapositive would be an easier route than a direct proof?

 

[Mark44]

Would it be because both powers of x are odd, which will mean it is always negative?

To be clearer, say what "it" refers to.

So proving the contapositive would be an easier route than a direct proof?

 

[tehdiddulator]

Still not entirely sure if I did it correctly...this class is giving me such a headache. Not sure if its the book or just my lack of understanding but some of these topics covered are going straight over my head...



Let x$\in$ℝ. If 3x$^{4}$+1 ≤ x$^{7}$ + x$^{3}$, then x > 0.
Proof
Contrapositive.
If x≤ 0, then 3x$^{4}$+1> x$^{7}$ + x$^{3}$
Assume x ≤ 0 for x $\in$ ℝ.
Since for all values of x less than or equal to zero would produce a true statement.
Since 3x$^{4}$+1 < 0 for x ≤ 0 and 0 > x$^{7}$ + x$^{3}$ for x ≤ 0.


Would this suffice as a proof? It seems like I don't know the tools to know how to get there, it always seems obvious when I look at the answer key, but have really no idea how to get there.

 

[tehdiddulator]

Oh...and I would just like to point out how the book does this type of problem. Seems incredibly non-intuitive to me.

Let x $\in$ ℝ. If x$^{5}$ -3x$^{4}$+2x$^{3}$-x$^{2}$ +4x - 1 ≥ 0, then x ≥ 0.
Proof.
Assume that x < 0. Then x$^{5}$ < 0, 2x$^{3}$ < 0, and 4x < 0. In addition, -3x$^{4}$ < 0 and -x$^{2}$ < 0.

Thus x$^{5}$ -3x$^{4}$+2x$^{3}$-x$^{2}$ +4x - 1 < 0 - 1 < 0.
as desired.

This is what I have to go on for an example...there has got to be a better way to do these proofs. I'm just regurgitating the same problem, except with different numbers and yet I have no idea what I'm doing.

 

[Mark44]

Still not entirely sure if I did it correctly...this class is giving me such a headache. Not sure if its the book or just my lack of understanding but some of these topics covered are going straight over my head...
Let x$\in$ℝ. If 3x$^{4}$+1 ≤ x$^{7}$ + x$^{3}$, then x > 0.
Proof
Contrapositive.
If x≤ 0, then 3x$^{4}$+1> x$^{7}$ + x$^{3}$
Assume x ≤ 0 for x $\in$ ℝ.
Since for all values of x less than or equal to zero would produce a true statement.

You can omit the sentence above. Instead, show using inequalities why this is so.

For any real x, 3x⁴ + 1 ≥ 1 > 0
For any x ≤ 0, x⁷ ≤ 0 and x³ ≤ 0, hence x⁷ +x³ ≤ 0
Then for any x ≤ 0, 3x⁴ + 1 > x⁷ +x³.

Since 3x$^{4}$+1 < 0 for x ≤ 0 and 0 > x$^{7}$ + x$^{3}$ for x ≤ 0.

This shows that the contrapositive is a true statement, and the contrapositive is equivalent to the original statement, so it is true as well.Would this suffice as a proof? It seems like I don't know the tools to know how to get there, it always seems obvious when I look at the answer key, but have really no idea how to get there.

 

[clamtrox]

Would it be because both powers of x are odd, which will mean it is always negative? So proving the contapositive would be an easier route than a direct proof?

Yes I think this is one of the most obvious cases where indirect proof is easier than direct proof. For a direct proof, you'd need to solve the equation 1+3x⁴-x³-x⁷=0.