Show that $(1+x)^{1+x} \ge 1 + x + x^2 + \frac{x^3}{2}$ for $x \ge -1$

Question

Show that for every $x\ge -1$ we have the inequality

$$(1+x)^{1+x} \ge 1 + x + x^2 + \frac{x^3}{2}$$ with equality only if $x = 0$.

Notes:

1. One of Bernoulli's inequalities : $(1+x)^a \ge 1 + a x$ if $x\ge -1$ and $a\ge 1$, so $(1+x)^{1+x}\ge 1 + x + x^2$ for $x \ge 0$.

2. We can further improve it as follows $(1+x)^{1+x} \ge 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \frac{x^5}{12} $ for $x \ge -1$.

3. One could look at the Taylor expansion of $(1+x)^{1+x}= \exp( (1+x) \log (1+x))$

Any feedback would be appreciated!

Answer 1

For any real $a$, we use the Taylor expansion of $(1+x)^a$ about $x = 0$, with Lagrange form of the remainder, and obtain $$(1+x)^a = 1 + ax + {a(a-1) \over 2}(1 + c)^{a -2} x^2$$ here $c$ is in the open interval $(0,x)$ or $(x,0)$. Substituting $a = 1 + x$ gives $$(1 + x)^{1 + x} = 1 + x + x^2 + {(1 + x)x \over 2}(1 + c)^{x - 1}x^2$$ Thus we need to show that ${(1 + x)x \over 2}(1 + c)^{x - 1}x^2 > {x^3 \over 2}$ when $x \neq 0$. Thus if $x > 0$ we must show $(1 + x)(1 + c)^{x-1} > 1$, and if $x < 0$ we must show $(1 + x)(1 + c)^{x-1} < 1$. We do this in three cases.

Case 1: $x \geq 1$. Then $(1 + x)(1 + c)^{x-1} \geq (1 + x)(1 + 0)^{x - 1} = 1 + x > 1$.

Case 2: $0 < x < 1$. Then $(1 + x)(1 + c)^{x-1} > (1 + x)(1 + x)^{x-1}$ since the exponent $x-1$ is negative and $c < x$. Then $(1 + x)(1 + x)^{x-1} = (1 + x)^x > (1 + 0)^x = 1$.

Case 3: $-1 < x < 0$. Then $(1 + x)(1 + c)^{x-1} < (1 + x)(1 + c)^{-1} < (1 + x)(1 + x)^{-1} = 1$

Answer 2

Let $x>-1$. Easy to show that $1+x+x^2+\frac{x^3}{2}>0$.

Let $f(x)=(1+x)\ln(1+x)-\ln\left(1+x+x^2+\frac{x^3}{2}\right)$ for $x>-1$.

Thus, $$f'(x)=\ln(1+x)+\frac{x^3-x^2-2x}{x^3+2x^2+2x+2},$$ $$f''(x)=\frac{x^2(x^4+7x^3+19x^2+28x+16)}{(x^3+2x^2+2x+2)^2(1+x)}\geq0$$ because for $x>-1$ $$x^4+7x^3+19x^2+28x+16>x^4+7x^3+19x^2+28x+15=$$ $$=(x+1)(x^3+6x^2+13x+15)=(x+1)((x+2)^3+x+7)>0$$ Thus, $f'$ increases for $x>-1$.

But $f'(0)=0$, which gives $$f(x)\geq f(0)=0.$$

Answer 3

In a previous post I show without derivative that :

$$\left(1+x\right)^{\left(1+x\right)}\geq \left(1-\sqrt{1+x}+1+x\right)^{\left(x+2\right)}$$

Now we can use Binomial theorem at second order as we have :

$$\left(1+\left(x+2\right)\left(1+x-\sqrt{x+1}\right)+\frac{1}{2}\left(x+1\right)\left(x+2\right)\left(1+x-\sqrt{x+1}\right)^{2}\right)\leq \left(1-\sqrt{1+x}+1+x\right)^{x+2}$$

Now as kids plays we can show :

$$\left(1+\left(x+2\right)\left(1+x-\sqrt{x+1}\right)+\frac{1}{2}\left(x+1\right)\left(x+2\right)\left(1+x-\sqrt{x+1}\right)^{2}\right)\geq 1+x+x^{2}+\frac{x^{3}}{2}$$

Performing $y^2-1=x$ and factoring .

Ps: We don't need derivative .

Answer 4

We have :

Let $x\geq 1$ then we have :

$$e^x\geq 1+x+x^2/2+x^3/6$$

Then setting $x\to x\ln x$ we have :

$$x^x\geq 1+x\ln\left(x\right)+\frac{x^{2}\ln^{2}x}{2}+\frac{\left(x\ln x\right)^{3}}{6}$$

Using for $x\geq 1$ :

$$\ln x \geq 1-1/x,\ln x\geq 2(x-1)/(x+1)$$

We have :

$$x^{x}\geq 1+x\ln\left(x\right)+\frac{\left(x\cdot\frac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right)^{2}}{2}+\frac{\left(2x\left(1-\frac{1}{\sqrt{x}}\right)\right)^{3}}{6}$$

Remains to show :

$$x+\left(x-1\right)^{2}+\frac{\left(x-1\right)^{3}}{2}<1+x\ln\left(x\right)+\frac{\left(x\cdot\frac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right)^{2}}{2}+\frac{\left(2x\left(1-\frac{1}{\sqrt{x}}\right)\right)^{3}}{6}$$

Now we have the intermediate inequality $x\geq 1$:

$1+x\ln\left(x\right)-x\geq \frac{\left(x-1\right)^{2}}{1+x}+\frac{1}{6}\frac{\left(x-1\right)^{3}}{1+x^{3}}\geq \frac{\left(x-1\right)^{3}}{2}+\left(x-1\right)^{2}-\left(\frac{\left(x\cdot\frac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right)^{2}}{2}+\frac{\left(2x\left(1-\frac{1}{\sqrt{x}}\right)\right)^{3}}{6}\right)$

Dividing by $x$ and introudcing :

$$f\left(x\right)=\frac{1}{x}+\ln\left(x\right)-1-\frac{1}{x}\frac{\left(x-1\right)^{2}}{1+x}-\frac{1}{6x}\frac{\left(x-1\right)^{3}}{1+x^{3}}$$

Then :

$$f'(x)=\frac{\left(x-1\right)^{3}\left(6x^{4}-5x^{3}+9x^{2}-3x+1\right)}{6x^{2}\left(x+1\right)^{2}\left(x^{2}-x+1\right)^{2}}$$

So the minimum is at $x=1$ so one of the side is shown .

For the other side we have $x\geq 1$:

$$\frac{\left(x^{2}-1\right)^{2}}{1+x^{2}}+\frac{1}{6}\frac{\left(x^{2}-1\right)^{3}}{1+x^{6}}-\left(\frac{\left(x^{2}-1\right)^{3}}{2}+\left(x^{2}-1\right)^{2}-\left(\frac{\left(x^{2}\cdot\frac{4\left(x-1\right)}{x+1}\right)^{2}}{2}+\frac{\left(2x^{2}\left(1-\frac{1}{x}\right)\right)^{3}}{6}\right)\right)\geq 0$$

Because :

$$\frac{(x-1)^{4}\left(5x^{10}+6x^{9}+26x^{8}+14x^{7}-13x^{6}-28x^{5}-10x^{4}+8x^{3}+26x^{2}+12x+2\right)}{6(x+1)^{2}(x^{2}+1)(x^{4}-x^{2}+1)}\geq 0$$

So we have shown the inequality for $x\geq 1$

Perhaps a similar path works for $0\leq x\leq 1$