How to prove $2e^x>x^3+3x$ holds for all $x \in \mathbb{R}$?

Question

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.

Answer 1

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.

Answer 2

Let $f(x)=2e^x-x^3-3x$ define a function $f:\mathbf R\to\mathbf R.$ Then $f'(x)=2e^x-3x^2-3,f''(x)=2e^x-6x$ and $f'''(x)=2e^x-6$; we then have that $f'^{v}(x)=2e^x>0\,\,\forall\, x.$ Thus, $f''$ is entirely convex. It is most minimal where $f'''(x)=0,$ which gives $x=\log3.$ Thus, $f''\left(\log3 \right)=6\left(1-\log3\right)<0$ since clearly, $e<3$ and $\log$ increases as $x\to\infty.$ Therefore, $f''$ has exactly two roots.

Now, $f''(0)=2>0$ and $f''(1)=2(e-3)<0.$ Thus, one root $r$ of $f''$ is in $(0,1).$ Also, $f''(2)=2(e^2-6)>0$ since $e^2=(5/2+\varepsilon)^2>15/2,$ where $1/6<\varepsilon<1.$ Thus, the other root $s$ of $f''$ is in $(1,2).$ Hence, both $r>0$ and $s>0.$ We then conclude that $f$ is convex in $(-\infty,r),$ concave in $(r,s)$ and convex again in $(s,\infty),$ with inflexions at $r$ and $s.$

When $x<0,$ it is easy to see that $f>0$; so we consider $f$ only for $x\ge0.$ Split this domain into $[0,2]$ and $(2,\infty).$ Then note that as $x\to2^+, f'\to 2e^2-15>0$ (by the reasoning about $e^2$ above). Also, since we know that $f''>0\,\,\forall\, x>s,$ it follows that $f'$ increases in $(2,\infty),$ since $2>s.$ Therefore $f'>0\,\,\forall\, x>2.$ It then follows that $f$ increases in $(2,\infty).$ Since as $x\to 2^+, f\to 2e^2-14>0$ (again by the reasoning about $e^2$ above), we have that $f>0\,\,\forall\, x\in(2,\infty).$

We now consider $f$ in $[0,2]=I.$ It is bounded there since it is entirely continuous. Hence, it attains its minimum value either where $f'=0$ or at the endpoints of $I.$ We show that whatever this minimum value is, it is positive, so that (conjoined with the demonstration above that $f>0$ for $x>2$) whenever $x\ge0,$ we must have that $f>0.$

Now when $x=0,$ we have $f(x)=2>0.$ At $x=2,$ we have $f(x)=2e^2-14>0$ again. So assume the minimum occurs when $f'=0$ instead (since otherwise the proof is complete). The function $f'$ increases in $(-\infty,r),$ decreases in $(r,s)$ and increases again in $(s,\infty),$ with turning points at $r$ and $s$ (all of this follows from what we established about $f''$ above). Since we also know that $r>0$ and $s>0,$ it follows that at most one root of $f'$ can be negative. Since $f'$ increases in $(-\infty,r)$ and $r>0,$ it must be the case that $f'$ has a $y$-intercept in $(-\infty,r).$ Indeed at $x=0, f'(x)=2-3<0.$ It follows that none of the roots of $f'$ is $0.$ Furthermore, we see that $f'$ therefore has no root in $(-\infty,0),$ so that all of its roots are positive. The proof is now complete.

Answer 3

Let $f(x)=(x^3+3x)e^{-x}$. Then $f(0)=0$ and $f(x)\to0$ as $x\to\infty$, so $f$ has a maximum at the point $x$ where $f'(x)=0$. This happens where $3x^2+3=x^3+3x$, i.e., at the solution to

$$2=x^3-3x^2+3x-1=(x-1)^3$$

which is to say, at $x=1+\sqrt[3]2$. Calculation (done by computer) says

$$f(1+\sqrt[3]2)=((1+\sqrt[3]2)^3+3(1+\sqrt[3]2))e^{-(1+\sqrt[3]2)}\approx1.9120322994\lt2$$

Or do you need to see that calulation done by hand?