So I was trying to prove that the sum of two log convex functions is also log convex. In my attemp I was able to conclude that if
$$f(x)^{1-t} f(y)^t + g(x)^{1-t} g(y)^t \leq (f(x)+g(x))^{1-t} (f(y)+g(y))^t $$
holds for any $t \in [0,1]$ then the proof is completed. I found that another user has asked how to prove the same question as me. This is the question How to prove that the sum of two log-convex functions is log-convex?
The first answer by the user Xiang Yu gets to the same point as me, and then he assume that $f(x)+g(x)=f(y)+g(y)=1$ and then continues the proof. Now I can´t figure out why we can assume this. I was trying to prove that this is indeed the worst case for the inequality to hold, but I am not sure this is true.
Can anybody explain me why we can assume $f(x)+g(x)=f(y)+g(y)=1$ , or maybe tell me what other options are to conclude the proof.
Thank you