Theorem. Let p∈(1,2]p \in (1, 2]. Then for x,y∈Rx, y \in \mathbb{R}, it holds that

12∣x−y∣p+12∣x+y∣p≤∣x∣p+∣y∣p. \frac{1}{2} |x - y|^p + \frac{1}{2} |x + y|^p \le |x|^p + |y|^p.

Proof. Let z=(a,b)∈R2z = \big(a, b\big) \in \mathbb{R}^2. Since p≤2p \le 2, it holds that

(∣a∣p+∣b∣p)1/p=∥z∥p≥∥z∥2=(∣a∣2+∣b∣2)1/2=(12∣a+b∣2+12∣a−b∣2)1/2. \left( |a|^p + |b|^p \right)^{1 / p} = \Vert z \Vert_p \ge \Vert z \Vert_2 = \left( |a|^2 + |b|^2 \right)^{1 / 2} = \left( \frac{1}{2}|a + b|^2 + \frac{1}{2}|a - b|^2 \right)^{1 / 2}.

Raising the power of both hand sides by pp, and via concavity of the function y=xp/2y = x^{p / 2}, we get

∣a∣p+∣b∣p≥(12∣a+b∣2+12∣a−b∣2)p/2≥12(∣a+b∣2)p/2+12(∣a−b∣2)p/2≥12∣a+b∣p+12∣a−b∣p. \begin{aligned} |a|^p + |b|^p &\ge \left( \frac{1}{2}|a + b|^2 + \frac{1}{2}|a - b|^2 \right)^{p / 2} \\ &\ge \frac{1}{2} \left( |a + b|^2 \right)^{p / 2} + \frac{1}{2} \left( |a - b|^2 \right)^{p / 2} \\ &\ge \frac{1}{2} |a + b|^p + \frac{1}{2} |a - b|^p. \end{aligned}

□\square

Remark: in general, for z∈Rnz \in \mathbb{R}^n, and 0<q≤p0 < q \le p, we have

∥z∥q=(∑i∈[n]∣zi∣q)1/q≥∥z∥p=(∑i∈[n]∣zi∣p)1/p \Vert z \Vert_q = \left( \sum_{i \in [n]} |z_i|^q \right)^{1 / q} \ge \Vert z \Vert_p = \left( \sum_{i \in [n]} |z_i|^p \right)^{1 / p}

To prove this, let yi=∣zi∣qy_i = |z_i|^q for each i∈[n]i \in [n], and r=p/q≥1r = p / q \ge 1. It suffices to show that

(∑i∈[n]yi)r≥∑i∈[n]yir. \left( \sum_{i \in [n]} y_i \right)^r \ge \sum_{i \in [n]} y_i^r.

Further, let xj=yj/(∑i∈[n]yi)x_j = y_j / \left( \sum_{i \in [n]} y_i \right) for each j∈[n]j \in [n]. It reduces to prove that

∑j∈[n]xjr≤∑j∈[n]xj=1. \sum_{j \in [n]} x_j^r \le \sum_{j \in [n]} x_j =1 .