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Let $a,b>0$, $f(x)=\dfrac{xa+(1-x)b}{a^xb^{1-x}},\quad x\in [0, 1].$ What is the maximum of $f(x)$?

I tried to find the derivative of $f(x)$, which is f'(x)=\left(\frac{a}{b}\right)^{1-x}\left(1-x\ln \frac{a}{b}\right)+\left(\frac{b}{a}\right)^{x}\left((1-x)\ln \frac{b}{a}-1\right). It seems difficult to find the critical points of $f(x)$.

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    The expression $(a/b)^xf'(x)$ being an affine function of $x$, to find the location of the maximum seems straightforward.2011-07-24

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It is natural to take the logarithm, but we do not have to. Here is an informative first step. Let $b=ca$. If we substitute $ca$ for $b$ in our expression, after a small amount of calculation we obtain $\frac{cx+(1-x)}{c^x}.$ This is useful, and not only as a simplifying device: It tells us that only the ratio $b/a$ matters.

Differentiation is straightforward: We get $\frac{c^x(c-1)-(cx+ 1-x)(\ln c)c^x}{c^{2x}}.$ The sign of the derivative is determined by the sign of $(c-1)-(cx+1-x)\ln c.$

Added: Suppose that $a \ne b$. We show that the maximum does not occur at an endpoint. By interchanging the roles of $a$ and $b$ if necessary, we can assume without loss of generality that $c>1$.

Let $g(x)=(c-1)-(cx+1-x)\ln c$. For $x$ positive but very close to $0$, $g(x)\approx -1-\ln c$. But from the Taylor expansion of $\ln(1+t)$, or otherwise, it is easy to see that $\ln(c) when $c-1$ is positive, so $g(x)$ is positive when $x$ is close enough to $0$. A similar argument shows that $g(x)$ is negative when $x$ is close enough to $1$. Thus the maximum occurs in the open interval $(0,1)$. In particular, the unique $x$ at which $(c-1)-(cx+1-x)\ln c=0$ is in the interior of $[0,1]$.

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    @Sunni: A bit of experimentation seems to show it is plausible. No proof $y$et!2011-07-25
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HINT: (Answering the original question: Where are the critical points of the function) It is enough to maximize $g(t)$, where $g(t) := \ln f(t) = \ln(at + (1-t)b) - t \ln a - (1-t) \ln b$

Now,

g'(t) = \frac{a-b}{at + (1-t)b} - \ln a + \ln b

and hence, g'(t) = 0, when

$\begin{align} \frac{a-b}{at + (1-t)b} = \ln (a/b) \end{align}$

(and so on...)

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    @Srivatsan: That's great with me!2011-07-24