I am interested in solving a rather peculiar inequality that, like some of my previous posts, is physics-related because it comes from a physical condition that must be satisfied by a mathematical derivation.
Consider $N$ different solid objects, each of mass $m_i \quad \forall i = 1, \ldots, N$ and heat capacity $c_i \quad \forall i = 1, \ldots, N$. Initially, the solids are each at temperature $T_i \quad \forall i = 1, \ldots, N$. The objects are brought into thermal contact and allowed to spontaneously (irreversibly) equilibrate until they reach a common equlibrium temperature $T_{\mathrm{equilibrium}}$, which turns out to be$$T_{\mathrm{equilibrium}}= \frac{\sum\limits_{i=1}^{n} m_i \ c_i \ T_i}{\sum\limits_{i=1}^{n} m_i \ c_i} := \frac{m_1 \ c_1 \ T_1 + \ldots + m_N \ c_N \ T_N}{m_1 \ c_1 + \ldots + m_N \ c_N}.$$After some calculations, the change in entropy of the system turns out to be$$\Delta S_{\mathrm{system}} = \sum\limits_{i=1}^{N} m_i \ c_i \ \ln \left[\frac{\sum\limits_{i=1}^{N} m_i \ c_i \ T_i}{T_i \sum\limits_{i=1}^{N} m_i \ c_i}\right].$$Assuming that the system is isolated from the outside, we have that the change in entropy for the environment is equal to $ \Delta S_{\mathrm{surroundings}} = 0$.We have $\require{cancel}$$$\Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{system}} + \cancelto{0}{\Delta S_{\mathrm{surroundings}}} \implies \Delta S_{\mathrm{universe}} = \Delta S_{\mathrm{system}}.$$So:$$\Delta S_{\mathrm{universe}} = \sum\limits_{i=1}^{N} m_i \ c_i \ \ln \left[\frac{\sum\limits_{i=1}^{N} m_i \ c_i \ T_i}{T_i \sum\limits_{i=1}^{N} m_i \ c_i}\right],$$and since the process is irreversible, we have $\Delta S_{\mathrm{universe}} > 0$, according to the entropic version of the 2nd Law of Thermodynamics.
So my goal is:
How to prove$$\bbox[5px,border:4px solid #C0A000]{\sum\limits_{i=1}^{N} m_i \ c_i \ \ln \left[\frac{\sum\limits_{i=1}^{N} m_i \ c_i \ T_i}{T_i \sum\limits_{i=1}^{N} m_i \ c_i}\right] > 0} \ ?$$
I only managed to solve it for $N=2$, and it was slightly involved.
$$\bigg(\Delta S_{\mathrm{universe}}\bigg)_{N=2} = \sum\limits_{i=1}^{2} m_i \ c_i \ \ln \left[\frac{\sum\limits_{i=1}^{2} m_i \ c_i \ T_i}{T_i \sum\limits_{i=1}^{2} m_i \ c_i}\right] = m_1 \ c_1 \ \ln \left[\frac{m_1 \ c_1 \ T_1 + m_2 \ c_2 \ T_2}{T_1 \ (m_1 \ c_1 + m_2 \ c_2)}\right] + m_2 \ c_2 \ \ln \left[\frac{m_1 \ c_1 \ T_1 + m_2 \ c_2 \ T_2}{T_2 \ (m_1 \ c_1 + m_2 \ c_2)}\right].$$ Let $C_i = m_i c_i$. So: $C_1 = m_1 c_1, \ C_2 = m_2 c_2$. Thus:
\begin{align} \Delta S_{\mathrm{universe}} &= C_1 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{T_1 \ (C_1 + C_2)}\right] + C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{T_2 \ (C_1 + C_2)}\right] \\ &= -C_1 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{T_1 \ (C_1 + C_2)}\right]^{-1} + C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{T_2 \ (C_1 + C_2)}\right] \\ &= C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{T_2 \ (C_1 + C_2)}\right] - C_1 \ln \left[\frac{T_1 \ (C_1 + C_2)}{C_1 \ T_1 + C_2 \ T_2}\right] \\ &= C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{C_1 \ T_2 + C_2 \ T_2}\right] - C_1 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_1}{C_1 \ T_1 + C_2 \ T_2}\right],\tag{$\star$} \end{align} so we have to prove
$$\Delta S_{\mathrm{universe}} = C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{C_1 \ T_2 + C_2 \ T_2}\right] - C_1 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_1}{C_1 \ T_1 + C_2 \ T_2}\right] > 0.$$
Let $$\alpha \equiv \frac{T_1 - T_2}{T_1}, \quad \beta \equiv \frac{C_1 + C_2}{C_2}.$$
- $\texttt{Condition 1)}$: body $1$ at temperature $T_1$ is the hottest and body $2$ at temperature $T_2$ is the coolest. So, $\min (T_1, T_2) = T_2, \ \max (T_1, T_2) = T_1$. Being $\min (T_1, T_2) < \max (T_1, T_2)$, we have: $\boxed{T_2 < T_1}.$ So:
$$T_2 < T_1 \iff \frac{T_2}{T_1} < 1 \iff - \frac{T_2}{T_1} > - 1 \iff 1 - \frac{T_2}{T_1} > 1 - 1 \iff \underbrace{\frac{T_1 - T_2}{T_1}}_{\displaystyle \alpha} > 0 \implies \boxed{\alpha > 0}.$$
- $\texttt{Condition 2)}$: by definition, $\boxed{C_1 > 0, \ C_2 > 0}.$ So:
$$\frac{C_1}{C_2} > 0 \iff \frac{C_1}{C_2} + 1 > 1 \iff \underbrace{\frac{C_1 + C_2}{C_2}}_{\displaystyle \beta} > 1 \implies \boxed{\beta > 1}.$$
- $\texttt{Condition 3)}$: by definition, $\boxed{T_1 > 0, \ T_2 > 0}.$ So:
$$\frac{T_2}{T_1} > 0 \iff - \frac{T_2}{T_1} < 0 \iff 1 - \frac{T_2}{T_1} < 1+ 0 \iff \underbrace{\frac{T_1 - T_2}{T_1}}_{\displaystyle \alpha} < 1 \implies \boxed{\alpha < 1}.$$
So: $$\boxed{0 < \alpha < 1} \quad \text{and} \quad \boxed{\beta > 1}.$$
From $($\star$)$, we have:
\begin{align*} \Delta S_{\mathrm{universe}} &= C_2 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{C_1 \ T_2 + C_2 \ T_2}\right] - C_1 \ln \left[\frac{C_1 \ T_1 + C_2 \ T_1}{C_1 \ T_1 + C_2 \ T_2}\right], \end{align*} from which \begin{align*} \frac{\Delta S_{\mathrm{universe}}}{C_2} &= \ln \left[\frac{C_1 \ T_1 + C_2 \ T_2}{C_1 \ T_2 + C_2 \ T_2}\right] - \frac{C_1}{C_2} \ln \left[\frac{C_1 \ T_1 + C_2 \ T_1}{C_1 \ T_1 + C_2 \ T_2}\right] \\ &= \ln \left[\dfrac{\frac{C_1}{C_2} \ T_1 + T_2}{\frac{C_1}{C_2} \ T_2 + T_2}\right] - \left(\frac{C_1}{C_2} \color{red}{+1 - 1}\right) \ln \left[\dfrac{\frac{C_1}{C_2} \ T_1 + T_1}{\frac{C_1}{C_2} \ T_1 + T_2}\right] \\ &= \ln \left[\dfrac{\frac{C_1}{C_2} + \frac{T_2}{T_1}}{\frac{C_1}{C_2} \ \frac{T_2}{T_1} + \frac{T_2}{T_1}}\right] - \left(\underbrace{\frac{C_1+C_2}{C_2}}_{\beta} - 1\right) \ln \left[\dfrac{\frac{C_1}{C_2} + 1}{\frac{C_1}{C_2} + \frac{T_2}{T_1}}\right] \\ &= \ln \left[\dfrac{\frac{C_1}{C_2} {\color{red}{+1}} + \frac{T_2}{T_1} {\color{red}{-1}}}{\left(\frac{C_1}{C_2} \color{red}{+1 - 1}\right) \ \frac{T_2}{T_1} + \frac{T_2}{T_1}}\right] - \left(\beta - 1 \right) \ln \left[\dfrac{\frac{C_1+C_2}{C_2}}{\frac{C_1}{C_2} {\color{red}{+1}} + \frac{T_2}{T_1} {\color{red}{-1}}}\right] \\ &= \ln \left[\dfrac{\frac{C_1+C_2}{C_2} + \frac{T_2-T_1}{T_1}}{\left(\frac{C_1+C_2}{C_2} -1\right) \ \frac{T_2}{T_1} + \frac{T_2}{T_1}}\right] - \left(\beta - 1\right) \ln \left[\dfrac{\beta}{\frac{C_1+C_2}{C_2} + \frac{T_2-T_1}{T_1}}\right] \\ &= \ln \left[\dfrac{\beta - \frac{T_1-T_2}{T_1}}{\left(\beta-1\right) \ \frac{T_2}{T_1} + \frac{T_2}{T_1}}\right] - \left(\beta - 1\right) \ln \left[\dfrac{\beta}{\beta - \frac{T_1-T_2}{T_1}}\right] \\ &= \ln \left[\dfrac{\beta - \alpha}{\beta \ \frac{T_2}{T_1} - \cancel{\frac{T_2}{T_1}} + \cancel{\frac{T_2}{T_1}}}\right] - \left(\beta - 1\right) \ln \left[\dfrac{\beta}{\beta - \alpha}\right] \\ &= \ln \left[\dfrac{\beta - \alpha}{\beta \ \left(\frac{T_2}{T_1} {\color{red}{-1 +1}}\right)}\right] - \beta \ln \left[\dfrac{\beta}{\beta - \alpha}\right] + \ln \left[\dfrac{\beta}{\beta - \alpha}\right] \\ &= \ln \left[\dfrac{\beta - \alpha}{\beta \ \left(\frac{T_2-T_1}{T_1} +1 \right)}\right] + \beta \ln \left[\dfrac{\beta - \alpha}{\beta}\right]- \ln \left[\dfrac{\beta-\alpha}{\beta}\right] \\ &= \ln \left[\dfrac{\beta - \alpha}{\beta \ \left(1 - \frac{T_1-T_2}{T_1} \right)}\right] + \beta \ln \left[1- \frac{\alpha}{\beta}\right] - \ln \left[1- \frac{\alpha}{\beta}\right] \\ &= \ln \left[\dfrac{\beta - \alpha}{\beta \ \left(1 - \alpha \right)}\right] + \beta \ln \left[1- \frac{\alpha}{\beta}\right] - \ln \left[1- \frac{\alpha}{\beta}\right] \\ &= \ln \left[\frac{1}{1-\alpha} \cdot \frac{\beta- \alpha}{\beta}\right]+ \beta \ln \left[1- \frac{\alpha}{\beta}\right]- \ln \left[1- \frac{\alpha}{\beta}\right] \\ &= \ln \left[\frac{1}{1-\alpha}\right] + \cancel{\ln \left[1 - \frac{\alpha}{\beta}\right]} + \beta \ln \left[1- \frac{\alpha}{\beta}\right] - \cancel{\ln \left[1- \frac{\alpha}{\beta}\right]} \\ &= - \ln (1- \alpha) + \beta \ln \left[1- \frac{\alpha}{\beta}\right]. \end{align*}
So:
$$\frac{\Delta S_{\mathrm{universe}}}{C_2} = \beta \ln \left[1- \frac{\alpha}{\beta}\right] - \ln (1- \alpha).$$
The function $\ln x$ is strictly concave $\forall x > 0$. Let us state the following
Definition. A continuous function $f(x)$ is strictly concave in an open interval $(a,b)$ when $$f(t \ a + (1-t) \ b) > t \ f(a) + (1-t) \ f(b) \qquad \forall t \in (0,1).$$
Since $T_2 < T_1 \iff \frac{T_2}{T_1}<1$ by hypotesis, having to choose $a < b$, for simplicity's sake let us assume $a = \frac{T_2}{T_1}$ and $b = 1$. We have:
$$b = 1 \qquad \text{and} \qquad a = \frac{T_2}{T_1} = \frac{{\color{blue}{T_1- T_1}} + T_2}{T_1} = \frac{T_1 - (T_1-T_2)}{T_1} = 1 - \underbrace{\frac{T_1-T_2}{T_1}}_{\displaystyle \alpha} = 1 - \alpha.$$
So, taking $a = 1 - \alpha, \ b = 1$ in the concavity equation:
\begin{align} \ln(t(1-\alpha) + 1- t) &> t \ln (1-\alpha) + (1-t) \underbrace{\ln 1}_{0} \\ \ln(\cancel{t} - \alpha \ t + 1- \cancel{t}) &> t \ln (1-\alpha), \end{align} leading to $$\ln (1 - \alpha \ t) > t \ln (1-\alpha).$$
Since $\beta > 1 \iff 0 < \frac{1}{\beta}<1$, having to choose $t \in (0,1)$, for simplicity's sake let us assume $t= \frac{1}{\beta}$. So we have:
\begin{align} \ln (1 - \alpha \ t) &> t \ln (1-\alpha) \\ \ln \left(1- \alpha \cdot \frac{1}{\beta}\right) &> \frac{1}{\beta} \ln (1-\alpha) \\ \beta \ln \left(1 - \frac{\alpha}{\beta}\right) &> \ln (1-\alpha) \qquad \text{(because $\beta > 0$)}. \end{align}
So:
$$\beta \ln \left(1 - \frac{\alpha}{\beta}\right) > \ln (1-\alpha) \iff \underbrace{\beta \ln \left(1 - \frac{\alpha}{\beta}\right) - \ln (1-\alpha)}_{\displaystyle \frac{\Delta S_{\mathrm{universe}}}{C_2}} > 0 \implies \frac{\Delta S_{\mathrm{universe}}}{C_2} > 0,$$ from which \begin{equation} \bbox[#FFFF88,12px]{\bigg(\Delta S_{\mathrm{universe}}\bigg)_{N=2}> 0} \qquad \blacksquare \end{equation}
Having proved the case $N=2$, is it possible to generalise to the generic case $N$, as I wrote earlier? How to prove that $\Delta S > 0$ also for generic $N$, taking into account the previous derivation for two bodies?