Alternate proof of $a^\alpha b^\beta \leq \alpha a + \beta b$ where $\alpha + \beta = 1$

Question

For $a,b > 0$ and $\alpha + \beta = 1$, show that $a^\alpha b^\beta \leq \alpha a + \beta b$.

I'm aware that this question was already posed on this site, for instance here. However I'm looking for an alternate proof. I read that this can also be proved using the concavity of the function $f(x) = x^\alpha (1-x)^\beta$. However I can't figure out the end of the proof.

Here's what I have so far:

First of all, it suffices to prove the inequality for $a+b = 1$. This because each term in the equality has degree 1 (since $\alpha + \beta = 1$). So the LHS of the inequality is $a^\alpha b^\beta = a^\alpha (1-a)^\beta = f(a)$.

Now $f'(x) = \alpha x^{\alpha - 1} (1-x)^\beta - \beta x^\alpha (1-x)^{\beta - 1}$. Since $f(0) = f(1) = 0$ and $f(x) > 0$ for $x \in (0,1)$, $f$ has a maximum on the interval $(0,1)$. To find this maximum set $f'(0) = 0$. This gives

$$ \alpha x^{\alpha - 1} (1-x)^\beta = \beta x^\alpha (1-x)^{\beta - 1}$$

which implies

$$ \alpha (1-x) = \beta x $$

so $x = \alpha$ is the maximum. So I'd expect $f(\alpha)$ to be equal to $\alpha a + \beta b$. But $f(\alpha) = \alpha^\alpha (1-\alpha)^{1-\alpha}$.

How do I arrive at the upper bound $\alpha a + \beta b$?

Answer 1

You want to prove the inequality $$f(x)=x^\alpha (1-x)^\beta \leq \alpha x + \beta (1-x)$$ for all $x$. Note that both sides depend on $x$, so finding the maximum value is not useful.

Instead, consider a point $0<x_0<1$. Now you can find the tangent to the graph of $f$ at the point $(x_0,f(x_0))$ : its slope is given by $$f'(x_0)=\alpha x_0^{\alpha-1} (1-x_0)^\beta-\beta x_0^\alpha (1-x_0)^{\beta-1}=f(x_0)\left(\frac{\alpha}{x_0}-\frac{\beta}{1-x_0}\right),$$ and it passes through the point $(x_0,f(x_0))$, so it must be given by the equation \begin{align} y & =f(x_0)\left(\frac{\alpha}{x_0}-\frac{\beta}{1-x_0}\right)(x-x_0)+f(x_0) .\end{align} In particular, if $x_0=\frac{1}{2}$, then we obtain the line $$y=(\alpha-\beta)\left(x-\frac{1}{2}\right)+\frac{1}{2}.$$ Because $f$ is concave, its graph must be under any tangent line; in particular we have $$f(x)\leq (\alpha-\beta)\left(x-\frac{1}{2}\right)+\frac{1}{2}=\alpha x -\beta x+\frac{1-\alpha+\beta}{2}=\alpha x -\beta x+\beta=\alpha x+\beta (1-x),$$ where in the last line we used the fact that $1-\alpha=\beta$.

Answer 2

For non-negatives $\alpha$ and $\beta$ it's just AM-GM: $$\alpha a+\beta b\geq a^{\alpha}b^{\beta}.$$

For $\alpha<0$ it's wrong. Try $a\rightarrow0^+$.

Answer 3

Note that $$a^\alpha b^\beta \leq \alpha a + \beta b$$ is equivalent to $$\alpha\ln a+\beta\ln b \leq \ln(\alpha a + \beta b).$$ WLG, let $a\le b$. Define $$ f(x)=\alpha\ln x+\beta\ln b - \ln(\alpha x + \beta b). $$ Then $$ f'(x)=\frac{\alpha}x-\frac{\alpha}{\alpha x + \beta b}=\frac{\alpha\beta (b-a)}{a(\alpha x+\beta b)}\ge0, $$ namely $f(x)$ is increasing in $[a,b]$. So $f(a)\le f(b)=0$ and hence $$\alpha\ln a+\beta\ln b \leq \ln(\alpha a + \beta b).$$