User:Jatrophalouvre/sandbox

Algebraic Proof
Let the functions be $$f(x), g(x), h(x)$$.

From the conditions, $$1+\frac{g}{f}+\frac{h}{f}=0$$. Taking the derivative and by the Quotient rule, we get $$\frac{g^\prime f-gf^\prime}{f^2}+\frac{h^\prime f-hf^\prime}{f^2}=0$$.

Multiplying by $$f^2$$ throughout, we get $$(g^\prime f-gf^\prime)+(h^\prime f-hf^\prime)=0$$.

Clearly, $$\gcd(g, g^\prime)\mid(g^\prime f-gf^\prime), \gcd(f, f^\prime)\mid(g^\prime f-gf^\prime), \gcd(h, h^\prime)\mid(h^\prime f-hf^\prime)$$. From $$(g^\prime f-gf^\prime)+(h^\prime f-hf^\prime)=0$$, we can see that $$\gcd(h, h^\prime)\mid(g^\prime f-gf^\prime)$$.

Since $$f(x), g(x), h(x)$$ are relatively prime, $$gcd(f, f^\prime), gcd(g, g^\prime), gcd(h, h^\prime)$$ are also relatively prime. Hence,$$gcd(f, f^\prime)\cdot gcd(g, g^\prime)\cdot gcd(h, h^\prime)\mid(g^\prime f-gf^\prime)$$.

If $$g^\prime f-gf^\prime=0$$, then $$\left(\frac{f}{g}\right)^\prime=0$$, implying that $$g\mid f$$. This contradicts with the fact that $$f, g, h$$ are relatively prime.

Hence, $$g^\prime f-gf^\prime\neq0$$, implying that $$gcd(f, f^\prime)\cdot gcd(g, g^\prime)\cdot gcd(h, h^\prime)\leq g^\prime f-gf^\prime$$.

$$\text{deg}(f, f^\prime)+ \text{deg}(g, g^\prime)+ \text{deg}(h, h^\prime) \leq \text{deg}(g^\prime f-gf^\prime)$$

$$\text{deg}f-n_0(f)+\text{deg}g-n_0(g)+\text{deg}h-n_0(h) \leq \text{deg}g+\text{deg}f-1$$, where $$n_0(f)$$ denotes the number of distinct roots of $$f(x)$$.

Cancelling out $$\text{deg}f$$ and $$\text{deg}g$$ and rearranging, we get $$\text{deg}h\leq n_0(fgh)-1$$.

Similarly, $$\text{deg}f\leq n_0(fgh)-1$$, $$\text{deg}g\leq n_0(fgh)-1$$. Hence, $$max\{\text{deg}f, \text{deg}g, \text{deg}h\}\leq n_0(fgh)-1$$, which was what we wanted to show.