Exists a uniformly convex norm on Banach space satisfying certain condition?

Lemma (Exercise 3.29, Brezis). Let $E$ be a normed vector space with a uniformly convex norm and fix $p > 1$. If $x$, $y \in \overline{N(0, M)} =: B$ are at least $\epsilon > 0$ apart, then there is some $\delta$ such that$$\left\|{{x + y}\over2}\right\|^p \le {{\|x\|^p + \|y\|^p}\over2} - \delta.$$

Suppose not for the sake of contradiction. Then there are sequences $(x_n)$ and $(y_n)$ in $B$ such that$$\|x_ - y_n\| > {{\|x\|^p + \|y\|^p}\over2} - {1\over n},$$and we may suppose that $|x_n| \to a$ and $y_n| \to b$ by compactness of $[0, M]$. Since$$a + b > \epsilon_0 \text{ and }{{a^p + b^p}\over2} \le \left({{a + b}\over2}\right)^p,$$we have that $a = b$. Let$$x_n' = {{x_n}\over{\|x_n\|}} \text{ and }y_n' = {{y_n}\over{\|y_n\|}}$$for which$${\epsilon\over a} + \text{o}(1) < \|x_n' - y_n'\| < 1 - \delta.$$Therefore,$$a^p + \text{o}(1) \le \left\|{{x_n + y_n}\over2}\right\|^p \le a^p(1 - \delta)^p + \text{o}(1),$$which is a contradiction.

Original Problem (Exercise 3.30, Brezis). If $E$ is a normed vector space that admits an equivalent uniform norm $|\cdot|_1$, then there is a uniformly convex norm $|\cdot|_2$ such that$$\|x\| \le |x|_2 \le k\|x\|$$for $k$ arbitrarily close to $1$.

Let$$|x|_2^2 = \|x\|^2 = \alpha|x_1|^2,$$then by the lemma,$$\|x\|^2 \le |x|_2 \le (1 + \alpha C^2)\|x\|^2,$$which can be made arbitrarily small and for any $x$, $y \in \overline{B(0, 1)}$ with respect to the new norm, then for any $|x - y|_2 > \epsilon$,$$\left|{{x + y}\over2}\right|_2^2 \le \left\|{{x + y}\over2}\right\|^2 + \left|{{x + y}\over2}\right|^2 \le \left\|{{x + y}\over2}\right\|^2 + {{|x|^2 + |y|^2}\over2} - \delta.$$

Only a partial answer:

If you take the norm $\|x\|'=\|x\|+\alpha |x|$ as gerw said in the comments, it is easy to see that since $\|.\|\sim |.|$, i.e $\exists \underline c,\overline c>0: \underline c\|x\|\leq |x|\leq \overline c \|x\|,\forall x\in E\,\,$ you will have $$\|x\|\leq \|x\|+\alpha|x|\leq (1+\alpha \overline c)\|x\|,\forall x\in E$$. It is left to prove that indeed $\|.\|'=\|.\|+\alpha|.|$ is uniformly convex. I do not know if this is true, but there is a weaker result in the book Metric Embeddings of Mikhail Ostrovskii in the end of section 9.3.1.

It states that the sum of a locally uniformly convex norm and a seminorm which is estimated from above by a multiple of this locally uniformly convex norm is a locally uniformly convex norm.

You can apply this to the (seminorm) $\|.\|$ and the (locally) uniformly convex norm $\alpha|.|$, where you have that $\|x\|\leq \frac{1}{\underline c}|x|=\frac{1}{\alpha \underline c}\alpha|x|$. So for a given $k>1$, find $\alpha>0$ small enough so that $1+\alpha\overline c\leq k$ and you get a locally uniformly convex norm $\|.\|'=\|.\|+\alpha|.|$ on $E$ such that $\|x\|\leq \|x\|'\leq k\|x\|,\forall x\in E$