$y_i \left[ 1 - \log \frac{t_i}{y_i} - \sum_{j=1}^C y_j \log \frac{t_j}{y_j}\right] = t_i \iff y_j=t_j \ \forall \ j=1,2,...,C$

Question

I'm trying to prove the following affirmation:

Given the constants $t_j \in (0,1)$ and the variables $y_j \in(0,1)$, for $j=1,2,...,C$: \begin{equation} y_i \left[ 1 - \log \frac{t_i}{y_i} - \sum_{j=1}^C y_j \log \frac{t_j}{y_j}\right] = t_i \iff y_j=t_j \ \forall \ j=1,2,...,C \end{equation}

I've made some simulations for differents values of $C$ and in all of them this is true, but I'd like to formally prove it or find a specific case in which the affirmation is false.

Thanks in advance to the community.

Answer 1

Equivalently, you have: $$1 - \log\dfrac{t_1}{y_1} - \dfrac{t_1}{y_1} = 1 - \log\dfrac{t_2}{y_2} - \dfrac{t_2}{y_2}=\dots 1 - \log\dfrac{t_C}{y_C} - \dfrac{t_C}{y_C} = \sum_{j=1}^C y_j\log\dfrac{t_j}{y_j}.$$

Since the function $f(x) = \log x + x$ is increasing and thus injective when $x>0,$ it must follow that $t_i = ay_i$ for some $a>0$ for all $i.$ Then your equation becomes: $$1-\log a - a = \log a(y_1+y_2+\dots +y_C).$$ If $a = 1$, we are done since we are trying prove that. If $0<a<1$ or $a = 1-x,\, 0<x<1$ then: $$1-\log a - a = x-\log(1-x) >0 ,$$ easily seen from first derivative test or the Taylor series. This would imply $y_1+y_2+\dots +y_c = \dfrac{1-\log a - a}{\log a} < 0,$ which is absurd. Similarly, if $a>1:$, $$y_1+y_2+\dots +y_c = \dfrac{1-\log a - a}{\log a} < \dfrac{1-a}{\log a} < 0.$$

Therefore, the only possibility is $a=1$ and thus $t_i = y_i.$

Answer 2

Calling $x_i = t_i/y_i$ we get

$$1-\log(x_i)-x_i = \sum_j y_j \log(x_j) $$

Because the RHS does not depend on $i$, and $f(x)=1-\log(x)-x$ is one to one, this implies $x_i=x$ is a constant.

Hence

$$ 1-\log(x)-x = \log(x) \sum y_j \implies 1 - a \log(x) -x = 0$$

with $a=1+\sum y_i >0$. Now, the function $1 - a \log(x) -x$ is continuous and decreasing on $(0,+\infty)$, hence the trivial root $x=1$ is the only one.

Hence $t_i=y_i$