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Let $a, b \in (0, \infty)$, $a. Prove that the equation $\left(\frac{a+b}{2}\right)^{x+y}=a^xb^y$ has at least one solution in $(a, b)$. Some suggestions? Thanks.

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    Hint: You can find a solution with $x+y=1$.2012-12-03

2 Answers 2

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This is a direct usage of the intermediate value theorem.

Let $f(t)=a^{1-t}b^t$. Then $f(t)$ is continuous and $f(0)=a$ and $f(1)=b$. Since $\frac{a+b}2$ is between $a$ and $b$, there must be a $t_0\in[0,1]$ such that $f(t_0)=\frac{a+b}2$ Let $x=1-t$ and $y=t$.

Indeed, we could just solve for $t_0$ by writing $f(t)=a\left(\frac{b}a\right)^t$ and yielding $t_0=\log_{b/a} \frac{a+b}{2a}$

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    Too awesome! Thanks!2012-12-03
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The equation can be solved for $x/y$ in terms of $b/a$ as $ \frac{x}{y} = \frac{\ln(b/a)}{\ln\left(\dfrac{1+b/a}{2}\right)} - 1 $

The statement that there exist $x,y \in (a,b)$ that satisfy the equation is equivalent to the assertion that for all $t > 1$,

$ 1/t \le \frac{\ln(t)}{\ln\left(\dfrac{1+t}{2}\right)} - 1 \le t $

Let $\eqalign{f(t) &= t \ln \left(\dfrac{1+t}{2}\right)\cr g(t) &= \ln(t) - \ln \left(\dfrac{1+t}{2}\right)\cr h(t) &= \dfrac{1}{t} \ln \left(\dfrac{1+t}{2}\right)\cr}$ Then we want to show that $f(t) > g(t) > h(t)$ for $t > 1$. As $t \to 0$, all three have limit $0$. So it suffices to show that $f'(t) - g'(t) > 0$ and $g'(t) > h'(t)$ for $t > 1$.

In fact we have $f'(t) - g'(t) = \ln\left(\dfrac{1+t}{2}\right) + 1 - \frac{1}{t} \ge 1 - \frac{1}{t} > 0$ $ g'(t) - h'(t) = \frac{1}{t^2} \ln\left(\dfrac{1+t}{2}\right) > 0$

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    thank your for your solution (+1)2012-12-03