Prove this inequality for all real x > 1. -- my own problem

[lookagain]

Let \(\displaystyle x\) be a positive real number, and \(\displaystyle x > 1.\)

Prove:

\(\displaystyle x^3 + x^2 > 5x - 3\)

 

[tkhunny]

Re: Prove this inequality for all real x > 1. -- my own prob

Can you use a picture? That should be easy.

What is it AT x = 1? >? <? =? How many intersections are there for x > 1?

 

[lookagain]

Re: Prove this inequality for all real x > 1. -- my own prob

No, don't try to use a picture for a proof. Use algebra. Also, pictures can be misleading.

 

[Deleted member 4993]

Re: Prove this inequality for all real x > 1. -- my own prob

\(\displaystyle (x^3+x^2)-(5x-3) \ = \ (x-1)^3 + 4(x-1)^2\)

Positive for x>1

typo fixed

 

[lookagain]

Re: Prove this inequality for all real x > 1. -- my own prob

\(\displaystyle (x^3-x^2)-(5x-3) \ = \ (x-1)^3 + 4(x-1)^2\)

Positive for x>1

I take the above to be closer to a typo/accident.

Your line should be the equivalent to this:

\(\displaystyle (x^3 + x^2) - (5x - 3) \ = \ (x - 1)^3 + 4(x - 1)^2\)

 

[Bob Brown MSEE]

No roots > 1

Another way to put it ...
(FTA)
A cubic with all 3 roots < 1+, will not have another root > 1
xxx+xx-5x+3 = (x+3)(x-1)(x-1)

 

[lookagain]

Another way to put it ...
(FTA)
A cubic with all 3 roots < 1+, will not have another root > 1
x^3 + x^2 - 5x + 3 = (x + 3)(x - 1)(x - 1)

Fixed at this point.