In a paper I am reading, there is a step that seems to come from the following inequality: $$(1+x)^\alpha \le 1+2^\alpha x,$$ where $0<x<1$. (Also, $3\le \alpha \le 9/2$ in the context of the paper, but the above probably holds for more general $\alpha$, say, $\alpha\ge 1$.) It is stated with no explanation, and I feel that there is a slick solution, but I am unable to prove it without calculus.

I was unsuccessful with the binomial expansion due to the generalized binomial coefficients.

I tried a calculus approach that assumes $0<x<1$ and $\alpha \ge 1$, and I think I was successful. Consider $f(x):=1+2^\alpha x - (1+x)^\alpha$, and note that $f(0)=0$ and $f(1)=1$. Then, $$f'(x)=2^\alpha-\alpha(1+x)^{\alpha-1}.$$ Since $f'(0)=2^\alpha-\alpha>0$, we know $f$ is increasing at $0$. If we find zero or one critical point in $[0,1]$, we are finished.

Setting $f'(x^*)=0$ gives \begin{align*} 2^\alpha &= \alpha(1+x^*)^{\alpha-1}\\ x^* &=\left(\frac{2^\alpha}{\alpha}\right)^{1/(\alpha-1)}-1 \ge 0\\ \end{align*} (because $2^\alpha>\alpha$), so we have one critical point in the positive reals, and we are finished.

I'm a little uncertain about these last few steps involving the critical point; is it correct?

Question: Is there an easier way to prove the inequality, and does it hold for more general $x$ and $\alpha$?


2 Answers 2


Consider $$ f(x)=1+(2^\alpha-1)x-(1+x)^\alpha\tag{1} $$ Note that $f(0)=f(1)=0$. The Mean Value Theorem says that for some $0\lt x_\alpha\lt1$, we have $f'(x_\alpha)=0$.

Furthermore, since $\alpha-1\ge0$, $f'(x)=(2^\alpha-1)-\alpha(1+x)^{\alpha-1}$ is non-increasing.

Thus, $f'(x_\alpha)\ge0$ for $0\le x\le x_\alpha$, and $f'(x_\alpha)\le0$ for $x_\alpha\le x\le1$. That is, $f(0)=0$, then $f(x)$ increases for $0\le x\le x_\alpha$, then $f(x)$ decreases for $x_\alpha\le x\le1$, then $f(1)=0$.

Therefore, $f(x)\ge0$ for $0\le x\le1$. That is, $$ 1+(2^\alpha-1)x\ge(1+x)^\alpha\tag{2} $$ which implies $$ 1+2^\alpha x\ge(1+x)^\alpha\tag{3} $$

Note that for $x\lt0$, the inequality fails, so $x\ge0$ is sharp. However, since $(2)$ is a bit stronger than $(3)$, at $x=1$, the left side of $(3)$ is $1$ greater than the right side. Thus, we can extend $x$ a bit beyond $1$, how far is determined by $\alpha$. For $\alpha=1$, we get that $$ 1+2x\ge1+x $$ which is true for all $x\ge0$. For $\alpha\gt1$, $(1+x)^\alpha$ grows faster than $1+2^\alpha x$ and so there will be some greatest $x$ where $(3)$ holds.


Here's a simple convexity argument. Let $g(x)=(1+x)^\alpha$, and note that $\alpha>1$ implies $g''(x)=\alpha(\alpha-1)(1+x)^{\alpha-2}>0$ for $0<x<1$; so $g(x)$ is concave up. In that case $gf(x)$ will be bounded above on the interval $[0,1]$ by $g(0)(1-x)+x g(1)$, i.e. the line segment connecting the endpoints of $g(x)$ on this interval. Since $g(0)=1$ and $g(1)=2^\alpha+1$, one has $g(0<x<1) \leq 1+(2^\alpha-1)x$. Note that this is a stronger bound than the one stated since $2^\alpha-1<2^\alpha$.

  • $\begingroup$ Nice, thanks! For other readers, note that $f(x)=(1+x)^\alpha$ in this answer is different from $f$ in my question. $\endgroup$
    – angryavian
    Jul 17, 2014 at 1:37
  • $\begingroup$ @angryavian: Good point. Probably best if I modify things slightly to match. $\endgroup$ Jul 17, 2014 at 1:47

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