How to prove the following inequality (log)

Question

How to Prove the following inequality: $$\ln { \sum _{ i }{ { e }^{ { { c }_{ i } } } } } \ge \frac { \sum _{ i }{ { c }_{ i }{ e }^{ { c }_{ i } } } }{ \sum _{ i }{ { e }^{ { c }_{ i } } } } ,\quad { c }_{ i }<0$$

Answer 1

The function $\phi:[0, \infty) \to \Bbb R$ defined by $$ \phi(t) = \begin{cases} 0 & \text{ for } t = 0 \\ -t \ln(t) & \text{ for } t > 0 \end{cases} $$ is concave ($\phi''(t) = -\frac 1t < 0$ for $t > 0$) with $\phi(0) \ge 0$, and therefore subadditive (compare Concave function – Properties). It follows that $$ \phi\left(\sum_{i=1}^n t_i \right) \le \sum_{i=1}^n \phi(t_i) $$ for any $t_1, \ldots, t_n \ge 0$.

Choosing $t_i = e^{c_i}$ gives: $$ - \left( \sum_{i=1}^n e^{c_i} \right) \cdot \ln \left(\sum_{i=1}^n e^{c_i} \right) \le \sum_{i=1}^n \left( - e^{c_i} c_i\right) $$ which is the desired inequality. The assumption $c_i < 0$ is not needed.

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Simpler solution, inspired by gimusi's answer: For each $1 \le i \le n$ $$ e^{c_i} \cdot \ln \left(\sum_{j=1}^n e^{c_j}\right) \ge e^{c_i} \cdot \ln e^{c_i} = c_i e^{c_i} \, . $$ Adding these inequalities gives the desired $$ \left( \sum_{i=1}^n e^{c_i} \right) \cdot \ln \left(\sum_{j=1}^n e^{c_j}\right) \ge \sum_{i=1}^n c_i e^{c_i} \, . $$ It is easy to see that the inequality is strict if $n \ge 2$.

Answer 2

Let indicate

$$c_i=\ln x_i \quad x_i\in(0,1)$$

thus

$$\ln { \sum _{ i }{ { e }^{ { { c }_{ i } } } } } \ge \frac { \sum _{ i }{ { c }_{ i }{ e }^{ { c }_{ i } } } }{ \sum _{ i }{ { e }^{ { c }_{ i } } } }\iff \ln \left(\sum_ix_i\right) \ge \frac{\sum_i \left(x_i \ln x_i\right)}{\sum_i x_i} \\\iff \ln \left[\left(\sum_ix_i\right)^{\sum_i x_i}\right] \ge \sum_i \ln x_i^{x_i} \iff e^{\ln \left[\left(\sum_ix_i\right)^{\sum_i x_i}\right]} \ge e^{\sum_i \ln x_i^{x_i}} \\\iff \left(\sum_ix_i\right)^{\sum_i x_i}\ge\prod_i x_i^{x_i} $$

which is true indeed

$$\left(\sum_ix_i\right)^{\sum_i x_i}=(x_1+x_2+...+x_n)^{(x_1+x_2+...+x_n)}= (x_1+x_2+...+x_n)^{x_1}\cdot(x_1+x_2+...+x_n)^{x_2}\cdot ...\cdot(x_1+x_2+...+x_n)^{x_n} \ge x_1^{x_1}\cdot x_2^{x_2}\cdot...\cdot x_n^{x_n}= \prod_i x_i^{x_i}$$

Note: the hypothesis that $c_i<0$ is unnecessary