Let $X$ be uniformly convex Banach space under both the norms $\|\cdot\|_1$ and $\|\cdot\|_2$. Consider the space $X$ under the norm $\|\cdot\|_X$ defined by $$ \|x\|_X=\|x\|_1+\|x\|_2. $$ My question is the space $X$ uniformly convex under the new norm $\|\cdot\|_X$? I think this can be proved in a similar way as in the proof given in: Uniform convexity of equivalent intersection norm
But I could not proceed with the last step there as mentioned: $$ \|\frac{x+y}{2}\|_{E\cap F}^2=\|\frac{x+y}{2}\|_{E}^2+\|\frac{x+y}{2}\|_{F}^2\leq 1-\delta-\delta' $$ Indeed with the above idea, here I could get that $$ \|\frac{x+y}{2}\|_X^{2}=(\|\frac{x+y}{2}\|_1+\|\frac{x+y}{2}\|_2)^{2}\leq 2\Big(\|\frac{x+y}{2}\|_1^{2}+\|\frac{x+y}{2}\|_2^{2}\Big) \leq 2(1-\delta-\delta'), $$ which is different than what we need. Can someone please help on how to proceed. Thanks in advance.