<s>[INST] Prove (or disprove): $a\times b=c\times d$, the solutions for $x$ in the equation $\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$ is only $\pm 1$. Prove (or disprove): If $a,b,c,d$ are positive real numbers with $a\times b=c\times d$, then the only solutions for $x$ in the equation $$\frac {a^x+b^x}{c^x+d^x} = \frac{a+b}{c+d}$$ are $x = \pm 1$. 
Other than the obvious $a=b=c=d$ solution.
 [/INST] Clearly, if $a = c, b = d$ or $a = d, b = c$, then all $x$ satisfy the equation. Hence the statement is false. However, one can show that this is the only exception, namely, if $a,b,c,d > 0$ such that $ab = cd$ and $\{a,b\}\neq\{c,d\}$, then $x = \pm 1$ is the only solution to $$\frac{a^x+b^x}{a+b} = \frac{c^x+d^x}{c+d}.$$
Suppose $ab = cd = s^2$. Without loss of generality, we may assume $s = 1$ since otherwise we can divide $a,b,c,d$ by $s$ and the solution set will not change. We may also assume $a \ge b$ and $c \ge d$. Now $b = 1/a, d = 1/c$ and $a,c \ge 1$.
Consider $f\colon [0,\infty)\times \mathbb{R}\to\mathbb{R}$ defined by $$f(z, x) = \frac{e^{zx} + e^{-zx}}{e^{z} + e^{-z}}.$$
Claim $f(z,x)$ is decreasing in $z$ if $|x| < 1$ and is increasing in $z$ if $|x| > 1$. Therefore $f$ is injective in $z$ if $x\neq \pm 1$.
Suppose the claim holds. We know that $f(\log a,x)=f(\log c,x)$ implies either $x = \pm 1$ or $a = c$.
Proof of Claim Observe that $$\partial_z f(z,x) = (-1 + x)\left(e^{(1+x)z}-e^{-(1+x)z}\right) + (1+x)\left(e^{(x-1)z}-e^{(1-x)z}\right).$$
If $|x| < 1$, then all terms above are negative and so $f$ is decreasing in $z$. If $|x| > 1$, then all terms above are positive, and so $f$ is increasing in $z$.
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