A Clarkson-type inequality $|a+b+c|^q + \sum\limits_{\mathrm{cyc}} |a + b - c|^q \ge \sum\limits_{\mathrm{cyc}} (|a + b|^q + |a - b|^q)$

Question

Show that, for $a,b,c \in \mathbb{R}$ and $q \geq 2$: $$ |a+b|^{q} + |a-b|^{q} + |a+c|^q + |a-c|^q + |b+c|^q + |b-c|^q \\\leq \\|a+b+c|^q + |a+b-c|^q + |a-b+c|^q + |a-b-c|^q $$

When $c=0$, this follows from the result here.

Is some generalization of Clarkson's inequality in the real case for more than 2 variables known?

Edit: Numerical experiments indicate that $\frac{lhs}{rhs}$ is maximized when one of $a, b, c$ is non-zero and the other two are zero. Many inequalities involving convex functions (for example: for $p\geq 1$, $(x_1^2 + x_2^2 + ... +x_n^2)^p \geq ( (x_1^2)^{p} + (x_2^2)^{p} + ... +(x_n^2)^{p})$ satisfy this property (the inequality being tight when one variable is non-zero and rest all is zero). This gives the impression that the inequality in question follows from some convexity arguments.

Answer 1

WLOG, assume that $a\ge b \ge c \ge 0$.

Fact 1: It holds that \begin{align*} &(a+b-c)^q + (a-b+c)^q + |b+c-a|^q \\ \ge\,& a^q + b^q + c^q + (a - b)^q + (b - c)^q + (a - c)^q. \end{align*} (The proof is given at the end.)

Fact 2: It holds that $$(a + b + c)^q + a^q + b^q + c^q \ge (a + b)^q + (b + c)^q + (c + a)^q.$$ (Note: Take derivative.)

Using Facts 1-2, the desired result follows.

---

Proof of Fact 1:

Let \begin{align*} &x_1 = (a + b - c)^2, ~ x_2 = (a - b + c)^2, ~ x_3 = (b + c - a)^2,\\ &y_1 = a^2, ~ y_2 = b^2, ~ y_3 = c^2, ~ y_4 = (a-b)^2, ~ y_5 = (b-c)^2, ~ y_6 = (a-c)^2. \end{align*} Let $p = (x_1, x_2, x_3, 0, 0, 0)$ and $q = (y_1, y_2, y_3, y_4, y_5, y_6)$.

First, we have $x_1 \ge x_2 \ge x_3$ and $y_1 = \max(y_1, y_2, y_3, y_4, y_5, y_6)$ and $y_2 \ge y_3$ and $y_6 - y_4 \ge 0$ and $y_6 - y_5 \ge 0$ and \begin{align*} x_1 - y_1 &\ge 0,\\ x_1 + x_2 - y_1 - y_6 &= 2ac + 2b^2 - 4bc + c^2 \\ &= 2(a-b)c + 2b(b-c) + c^2\\ &\ge 0, \\ x_1 + x_2 - y_1 - y_2 &= a^2 + b^2 - 4bc + 2c^2\\ &= 2c(a - b) + 2(b-c)^2 + 2(a-b)^2 + 2(a-b)(b-c)\\ &\ge 0. \end{align*}

Second, we have $$x_1 + x_2 + x_3 = y_1 + y_2 + y_3 + y_4 + y_5 + y_6.$$

Thus, $q$ is majorized by $p$.

Note that $u\mapsto u^{q/2}$ is convex on $u \ge 0$. Using Karamata inequality, the desired result follows.