The equation $\left(\frac{a+b}{2}\right)^{x+y}=a^xb^y$ Let $a, b \in (0, \infty)$, $a<b$. 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.
 A: 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 $t_0=\log_{b/a} \frac{a+b}{2a}$
A: 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 $$f'(t) - g'(t) = \ln\left(\dfrac{1+t}{2}\right) + 1 - \frac{1}{t} \ge  1 - \frac{1}{t} > 0$$
$$ g'(t) - h'(t) = \frac{1}{t^2}  \ln\left(\dfrac{1+t}{2}\right) > 0$$
