Lower bound on smallest eigenvalue of symmetric matrix

Question

Let $A \in \mathbb{R}^{n \times n}_{++}$ a matrix with strictly positive eigenvalues denoted by $\lambda_1, \lambda_2, \dots, \lambda_n$. We know that

$$ \lambda_1 + \cdots + \lambda_n = \mbox{Tr}(A)$$

From section II of Waldron's paper$^\star$,

$$\color{blue}{\text{min} \{\lambda_1, \dots , \lambda_n \} \geq \frac{\lambda_1 + \cdots + \lambda_n}{n} =: \bar{\lambda}}$$

I do not know how to prove this, or modify it to be true. The most similar formula I can thing of right now is obtained using the Samuelson's inequality. In such a way I obtain: $$ \lambda_i \geq \bar{\lambda} - \sqrt{\frac{n-1}{n}} \cdot \sqrt{\sum_{k=1}^n (\lambda_i - \bar{\lambda})^2} $$

Below is a snip from the mentioned paper.

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_{$\star$ Waldron, S. (2003). Generalized welch bound equality sequences are tight frames. IEEE Transactions on Information Theory, 49(9), 2307–2309.}

Answer 1

The formula is wrong: the average is always bigger than the minimal unless the numbers are all equal.