proof

[BigBeachBanana]

what about x>0 then?

$$\text{For}\, x>0\Rightarrow 3(x^3-1)^2>0>-4x^5\Rightarrow3(x^3-1)^2>-4x^5 \Rightarrow 3(x^3-1)^2 + 4x^5>0$$

 

[BigBeachBanana]

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.

$$3(x^3 - 1)^2 > -4x^5$$

Nvm, my logic doesn't work for x<0

 

[canvas]

$$\text{For}\, x>0\Rightarrow 3(x^3-1)^2>0>-4x^5\Rightarrow3(x^3-1)^2>-4x^5 \Rightarrow 3(x^3-1)^2 + 4x^5>0$$

It's also not true, just put x=1

 

[canvas]

It's also not true, just put x=1

Ohh sorry, it is... My mistake it's too late, 3 a.m. my time, let's go back to this problem tomorrow...

 

[canvas]

$$\text{For}\, x>0\Rightarrow 3(x^3-1)^2>0>-4x^5\Rightarrow3(x^3-1)^2>-4x^5 \Rightarrow 3(x^3-1)^2 + 4x^5>0$$

but it should be:
$$\text{for}\,\,x>0:\,\,3(x^3-1)^2\red{\geq}0>-4x^5\Rightarrow3(x^3-1)^2+4x^5>0$$

 

[blamocur]

My turn . For $x > 0$ : $f(x) = x^6 +4x^5-6x^3+3 = 3(x^3-1)^2 + 4x^5 > 0$, and we know that $f(0) = 3 > 0$.
Now, it is enough to prove that $f^\prime(x) \leq 0$ for $x < 0$.
$\frac{1}{2x^2}f^\prime(x) = 9x^3+10x^2-9 = 9x(x+1)^2 - (8x^2+9x+9) = g(x) - h(x)$
But $g(x) \leq 0$ when $x \leq 0$, and $h(x)>0$ for all $x$.

 

[Cubist]

Can't see the way that it can help ;/

It helps because it leads to a simple and elegant proof for half the domain of x. Now you just have to prove it for x<0

EDIT: I see that @blamocur has provided that for you in post#26

 

[lookagain]

. For $x > 0$ : $f(x) = x^6 +4x^5-6x^3+3 = 3(x^3-1)^2 + 4x^5 > 0$,

You meant that $\displaystyle \ f(x) \ = \ 3x^6 + \ ... \$

 

[blamocur]

You meant that $\displaystyle \ f(x) \ = \ 3x^6 + \ ... \$

I sure did -- thank you for catching the typo.

 

[JeffM]

That's good reasoning, what about x>0 then? In case of x=0 is obvious

I start by repeating what I wrote in post # 12

$$y(x) = 3x^6+4x^5-6x^3+3 \implies y' = 18x^5 + 20x^4 - 18x^2.$$
$$a(x) = (3x^6 - 6x^3) \text { and } b(x) = 4x^5 + 3 \implies\\ a + b = y.$$
$$a' = 18x^5 - 18x^2 = 18x^2(x^3 - 1) \implies a > - 3 \text { if } x \ne 1 \text { and } a(1) = - 3.$$
$$b(0) = 3 \text { and } b > 3 \text { if } x > 0.$$
$$y(0) = 3 > 0, \ y(1) = 4 > 0.\\ 0 < x < 1 \implies b > 3 \text { and } a > - 3 \implies y = a + b > 0.\\ 1 < x \implies b > 7 \text { and } a > - 3 \implies a + b > 4 > 0.$$
$$\therefore x \ge 0 \implies y > 0.$$
$$b' = 20x^4 > 0 \text { if } x \ne 0 \text { and } b'(0) = 0 \implies\\ b \text { is monotonically increasing.}$$
$$x < 0 \implies a > 0. \\ b \left ( \left \{ - \dfrac{3}{4} \right \}^{1/5} \right ) = 0.\\ \left ( - \dfrac{3}{4} \right )^{1/5} \le x < 0 \implies y > 0.$$
Putting that together with the previous result, we have

$$\left ( - \dfrac{3}{4} \right )^{1/5} \le x \implies y > 0.$$
At this point we need a different decomposition of y.

$$c(x) = 4x^5 - 6x^3 = 2x^3(2x^2 - 3) \text { and } d(x) = 3x^6 + 3 \implies c + d = y.$$
$$x = - \sqrt{\dfrac{3}{2}} \implies c = 0 \text { and } - \sqrt{\dfrac{3}{2}} < x < 0 \implies c > 0.$$
$$\text {But } d \ge 3 \text { for all } x \implies y \ge 0 + 3 > 0 \text { if } - \sqrt{\dfrac{3}{2}} < x < 0.$$
Putting that together with the previous result, we have

$$- \sqrt{\dfrac{3}{2}} \le x \implies y > 0.$$
I am running out of time here. Do you see the last step? Give it a go. I'll be back in an hour or two.

 

[canvas]

After some hard work guys I found the best solution for that, just look:

$$3x^6+4x^5-6x^3+3=2x^2(x^2+x-1)^2+(x^3-1)^2+(x^2-1)^2+x^4+1\geq1>0\\\\Q.E.D.$$

 

[blamocur]

After some hard work guys I found the best solution for that, just look:

$$3x^6+4x^5-6x^3+3=2x^2(x^2+x-1)^2+(x^3-1)^2+(x^2-1)^2+x^4+1\geq1>0\\\\Q.E.D.$$

Perfect!

 

[canvas]

Perfect!

Thank you!

 

[Cubist]

You can obtain a tighter bound using this, rather obvious , equivalent...

$\displaystyle 3x^6 + 4x^5 - 6x^3 + 3$
$\displaystyle =\left( (4x^3 + 5x^2 - 6x)^2 + (6x^3 + 4x^2 - 5)^2 + (2x^3 - 2x^2)^2 + 4x^6 + 3x^4 + 4x^2 \right)/20 + \frac{7}{4}$
$\displaystyle \ge 1\frac{3}{4} > 0$

( It probably would have been quicker if I'd reinvented calculus from scratch rather than finding the above expression ? )

 

[Deleted member 4993]

You can obtain a tighter bound using this, rather obvious , equivalent...

$\displaystyle 3x^6 + 4x^5 - 6x^3 + 3$
$\displaystyle =\left( (4x^3 + 5x^2 - 6x)^2 + (6x^3 + 4x^2 - 5)^2 + (2x^3 - 2x^2)^2 + 4x^6 + 3x^4 + 4x^2 \right)/20 + \frac{7}{4}$
$\displaystyle \ge 1\frac{3}{4} > 0$

( It probably would have been quicker if I'd reinvented calculus from scratch rather than finding the above expression ? )

That is called perseverance...