Existence of solution in $x,y \in (a,b)$ of $(\frac { a+b}2)^{x+y}=a^xb^y$

Question

Let $a<b$ be positive real numbers , then is it true that there exist $x,y \in (a,b)$ such that

$ \bigg(\dfrac { a+b}2\bigg)^{x+y}=a^xb^y$ ?

Answer 1

Fix $a\lt b$ positive. The function $F$ defined by $F(u)=a^{1-u}b^{u}$ for every $u$ in $[0,1]$ increases from $F(0)=a$ to $F(1)=b$. By continuity, and since $c=\frac12(a+b)$ is in $(a,b)$, there exists some $u(c)$ in $(0,1)$ such that $F(u(c))=c$. Every $(x,y)$ such that $u(c)x=(1-u(c))y$ solves $c^{x+y}=a^xb^y$ hence it remains to see whether one such couple $(x,y)$ is such that $x$ and $y$ are in $(a,b)$.

This happens if and only if, when $x=a$ and $u(c)x=(1-u(c))y$ then $y\lt b$, that is, if and only if $u(c)a\lt(1-u(c))b$, or, equivalently, $u(c)\lt b/(a+b)$, or, equivalently, $c\lt F(b/(a+b))$, that is, $c^{a+b}\lt a^ab^b$, that is, finally, $(a+b)\log c-a\log a-b\log b\lt0$.

Fix some $a\gt0$ and, for every $b\geqslant a$, consider $G(b)=(a+b)\log\left(\frac12(a+b)\right)-a\log a-b\log b$, then $G(a)=0$ and $G'(b)=\log\left(\frac12(a+b)\right)-\log b\lt0$ for every $b\gt a$ hence $G(b)\lt0$ for every $b\gt a$, QED.

Answer 2

We can solve the equation by $$(x(t),y(t)) = \left(\ln\left(\frac{2b}{a+b}\right)t,\ln\left(\frac{a+b}{2a}\right)t\right)$$

Now $\frac{a+b}{2}\geq a$ and therefore $\ln\left(\frac{a+b}{2a}\right)\geq 0$ and by $\frac{a+b}{2}\leq b$ we conclude $\ln\left(\frac{2b}{a+b}\right)\geq 0$.

Furthermore $\ln\left(\frac{a+b}{2a}\right) \geq \ln\left(\frac{2b}{a+b}\right)$

Thus for $$t \in \left(\frac{a}{\ln\left(\frac{2b}{a+b}\right)},\frac{b}{\ln\left(\frac{a+b}{2a}\right)}\right)$$ we have $x(t),y(t)\in (a,b)$.

Answer 3

Another way: With $t=\frac{x}{x+y}$, we seek solutions for $$\frac{a+b}2=a^tb^{1-t}$$ With the RHS being continuous, for solutions to exist in the right interval, it is sufficient to show that $$a^\frac{b}{a+b}b^\frac{a}{a+b}<\frac{a+b}2<a^\frac{a}{a+b}b^\frac{b}{a+b}$$

The left inequality is a consequence of weighted AM-GM, the right one follows on taking logs and using Jensen's inequality.

Answer 4

There should be an infinite number of solutions.

If we let x = f (t), y = g (t) for two continuous functions on [0, 1] with f (0) = 1, g (0) = 0, and f (1) = 0, g (1) = 1, then the left hand side of the equation is (a + b) / 2 at t = 0 and t = 1, while the right hand side changes from a < (a + b)/2 continuously to b > (a + b)/2 when t goes from 0 to 1. So for some 0 < t < 1 both sides must be equal. And since we picked f, g with very little restrictions, there should be a huge set of solutions.

Oops. Just found the restriction a <= x, y <= b.