Let $1\leq p<\infty$ and we define $l_p$ space be the vectors $x= (x_1, x_2, \dots)$, such that
$$\|x\|_p = \left(\sum_{n=1}^\infty \|x_n|^p\right)^{1/p}<\infty$$
Find the minimum constant $C$ such that for all $t\in [0,1]$ and vectors $x, y \in l_p$ space, with $\|x\|_p = \|y\|_p$, the following inequality holds
$$\|x+y\|_p \leq C\|tx + (1-t)y\|_p$$
My try: First observation is $C\geq 2$ since we can let $x = y$ then we can easily get $\|2x\|_p\leq C \|x\|_p$.
I guess the minimum constant may be $2$. To prove for $C = 2$ the inequality holds we need to prove that $$\|\frac{x+y}{2}\|_p\leq \|tx + (1-t)y\|_p$$ Which is to prove the function $\varphi(t) = \|tx + (1-t)y\|_p$ takes minimum on $\varphi(1/2)$. However I'm stuck here.
I know that $\varphi(1) = \varphi(0)$ and $\varphi$ is convex function of $t$. So calculating $\left.\frac{\mathrm d}{\mathrm d t}\varphi\right|_{t = 1/2}$ may help but it does not work.
