<s>[INST] How to prove $2e^x>x^3+3x$ holds for all $x \in \mathbb{R}$? Problem
Prove that $$2e^x>x^3+3x, \forall x \in \mathbb{R}.$$
Notes
It's clear that the inequality holds for all $x \leq 0$. Hence, we only need to reserch the situation when $x>0.$
Let $f(x)=2e^x-x^3-3x,(x>0).$ Here, if you use Tayor's formula of $e^x$, namely,$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots,$$ and plug it into $f(x)$,you'll have $$f(x)=2-x+x^2-\frac{2}{3}x^3+\cdots.$$ This will perhaps give nothing helpful. If you consider applying the derivative, $$f'(x)=2e^x-3x^2-3.$$ You can hardly find its zero-point, though there exists one root for $f'(x)=0$,say,$x_0=2.0715\cdots$ such that $f(x)$ reaches its minimum value,say, $$f(x)\geq f(x_0)=0.76990\cdots>0.$$
It's not easy to obtain these results without computer. Is there another tricky solution for this? Thanks in advance.
 [/INST] For $x\leq0$ it's obvious.
But for $x>0$,
$$2e^x-x^3-3x
    >2\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)-x^3-3x=$$
   $$=\frac{1}{60}(x^5+5x^4-40x^3+60x^2-60x+120)=$$
$$=\frac{1}{60}\left(\left(x-\frac{5}{2}\right)^2\left(x^3+10x^2+\frac{15}{4}x\right)+\frac{5}{16}(52x^2-267x+384)\right)>0.$$
These coefficients we can get by the following way.
By calculator we can get that the polynomial $x^5+5x^4-40x^3+60x^2-60x+120$ for $x\geq0$ gets a minimal value around $x=\frac{5}{2}.$
Now, we'll choose $a$, $b$ and $c$ such that
$$x^5+5x^4-40x^3+60x^2-60x+120=\left(x-\frac{5}{2}\right)^2(x^3+ax^2+bx+c)+Ax^2+Bx+C,$$ $$x^3+ax^2+bx+c>0$$ and $$Ax^2+Bx+C\geq0$$ for $x\geq0.$
Indeed, we need $$x^5+5x^4-40x^3=\left(x^2-5x+\frac{25}{4}\right)(x^3+ax^2+bx+c)+Dx^2+Ex+F,$$
which gives the following system:
$$5=a-5,$$
$$-40=b-5a+\frac{25}{4},$$ which gives $a=10$ and $b=\frac{15}{4}.$
Now, we know that
$$x^5+5x^4-40x^3+60x^2-60x+120-\left(x-\tfrac{5}{2}\right)^2\left(x^3+10x^2+\tfrac{15}{4}x+c\right)=Ax^2+Bx+C$$
and it remains to choose a value of $c$, for which
$$Ax^2+Bx+C>0$$ and we see that $c=0$ is valid.
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