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canvas

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Prove that [math]\forall x\in\R\,\,(3x^6+4x^5-6x^3+3>0)[/math]
Any hint please? I don't need the solution but some hints to prove it! Thanks
 
Does it have a minimum or multiple minimums? How could you find it/them?
 
[math]\frac{d}{dx}3x^6+4x^5-6x^3+3=x^2(18x^3+20x^2-18)\\\\\text{but now} \,\, 18x^3+20x^2-18\,\,\text{is the problem...}[/math]
 
You could use Newton's method to solve the cubic. Won't take many iterations to get a workable estimate.
 
It's actually not that bad. You know that x = 0 gives a minimum, so you can compare f(0) to 0. Once you find the other minimum value, compare f(value) to 0. I grant that there may be a more formal way to do this proof, but this is the approach I would take.
 
canvas said:
Prove that [math]\forall x\in\R\,\,(3x^6+4x^5-6x^3+3>0)[/math]
Any hint please? I don't need the solution but some hints to prove it! Thanks
Click to expand...
Here's an idea by factoring out the highest even power and completing the square. Continue with the remaining powers. I'm not sure if actually work, but give it a try:
[math]3x^6+4x^5-6x^3+3=x^4(3x^2+4x)-6x^3+3\\ =x^4\left[3\left(x+\frac{2}{3}\right)^2-\frac{4}{3}\right]-6x^3+3=3x^4\left(x+\frac{2}{3}\right)^2-x^4\left(\frac{4}{3}\right)-6x^3+3[/math]
 
R.M. said:
It's actually not that bad. You know that x = 0 gives a minimum, so you can compare f(0) to 0. Once you find the other minimum value, compare f(value) to 0. I grant that there may be a more formal way to do this proof, but this is the approach I would take.
Click to expand...
Me too. No algebraic short-cut jumps out at me.
 
It's interesting that this problem comes from high school lvl where students don't know Newtons method, I'm confused...
 
canvas said:
It's interesting that this problem comes from high school lvl where students don't know Newtons method, I'm confused...
Click to expand...
Don't be. I am working on an answer that I think will work with just a bit of calculus.
 
canvas said:
Prove that [math]\forall x\in\R\,\,(3x^6+4x^5-6x^3+3>0)[/math]
Any hint please? I don't need the solution but some hints to prove it! Thanks
Click to expand...
[math]y(a) = 3x^6+4x^5-6x^3+3 = (3x^6- 6x^3) + (4x^5 + 3) = a(x) + b(x)[/math]
[math]a' = 18x^5 - 18x^2 = 18x^2(x^3 - 1) \implies a' = 0 \text { at } x = 0 \text { or } x = 1.[/math]
[math]a(1) = 4 * 1^5 - 6 * 1^5 = - 2 .[/math]
[math]x \ge 0 \implies b(x) \ge 3 \text { and } a(x) \ge -2 \implies\\ y(x) \ge 1 >0.[/math]
Now what?
 
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I haven't had time to read JeffM's post, but I'll share this thought in case it helps. If you rewrite the inequality as follows then you should be able to prove it for the cases x=0 and x>0 but I can't see an easy way for x<0.

[math]3(x^3 - 1)^2 > -4x^5[/math]
 
Cubist said:
I haven't had time to read JeffM's post, but I'll share this thought in case it helps. If you rewrite the inequality as follows then you should be able to prove it for the cases x=0 and x>0 but I can't see an easy way for x<0.

[math]3(x^3 - 1)^2 > -4x^5[/math]
Click to expand...
Can't see the way that it can help ;/
 
Cubist said:
I haven't had time to read JeffM's post, but I'll share this thought in case it helps. If you rewrite the inequality as follows then you should be able to prove it for the cases x=0 and x>0 but I can't see an easy way for x<0.

[math]3(x^3 - 1)^2 > -4x^5[/math]
Click to expand...
[math]\text{For}\, x<0, -4x^5>0 \Rightarrow 3(x^3-1)^2+(-4x^5)>0[/math]
 
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