Suppose that $\phi : \mathbb{R} \mapsto \mathbb{R}$ is the "log barrier penalty function" with some limit parameter $a>0$, which is defined as $$ \phi(u) \triangleq \begin{cases}-a^2\log(1-(\frac{u}{a})^2) & |u| < a\\ \infty & \text{otherwise}\end{cases} $$
Apparently it can be shown that if $u \in \mathbb{R}^N$ satisfies $||u||_{\infty} < a $ then $$ ||u||_2^2 \leq \sum_{i=1}^N\phi(u_i) \leq \frac{\phi(||u||_{\infty})}{||u||^2_{\infty}}||u||_2^2 $$
Certainly, if $||u||_{\infty}<a \implies \phi(u_i) = -a^2\log(1-(\frac{u_i}{a})^2) \ \forall i$ which also tells us that $$ \sum_{i=1}^N\phi(u_i) = -\sum_{i=1}^Na^2\log(1-(\frac{u_i}{a})^2) < \infty $$
And certainly $$ ||u||_2^2 < Na^2$$
How would one then proceed?
Since $\log(1-x) \leq -x$ for all $x > 0$, we have $$ \sum_{i=1}^n \phi(u_i) = -a^2 \sum_{i=1}^n \log\left(1 - (u_i/a)^2\right) \geq a^2 \sum_{i=1}^n (u_i / a)^2 = \Vert u \Vert^2_2\,. $$ For the other direction, are you missing a square on the $\Vert \cdot \Vert_\infty$ norm term? With this you have: $$ -a^2 \sum_{i=1}^n \log(1 - (u_i / a)^2) = \sum_{i=1}^n u_i^2 (-a^2/u_i^2) \log(1 - (u_i / a)^2) \leq \frac{\Vert u \Vert^2_2 \phi(\Vert u\Vert_\infty)}{\Vert u \Vert_\infty^2}\,, $$ where the last inequality follows because $x\mapsto -1/x^2 \log(1 - x^2)$ is increasing. Without the square on the $\Vert \cdot \Vert_\infty$ norm in the denominator the inequality would not be homogeneous in the way you would expect and should not be true.