Clarkson inequality for complex numbers Let $1<p<2$. I'm trying to prove the inequality
$$
|a+b|^q+|a-b|^q\leq 2\big( |a|^p + |b|^p \big)^{q-1}
$$
where $\frac{1}{p}+\frac{1}{q}=1$.
Following this paper, I am able to prove the inequality for real $a,b$.
I'm missing the step to complex.
I have tried to set $x=|a+b|^2$ and $y=|a-b|^2$ (and other combinations) and then use the real inequality for $x,y$ with no success.
This is not exactly a duplicate of
On the second Clarkson's inequality
because that one asked from a "from scratch" proof while I'm only asking for one step.
 A: Let's assume for simplicity that $a\neq 0$ and $b\neq 0.$  Then $$\displaylines{|a+b|^2=|a|^2+|b^2|+2\Re (a\bar b)=
|a|^2+|b^2|+2|a|\,|b|\cos t\\ |a-b|^2=|a|^2+|b^2|-2\Re (a\bar b)=
|a|^2+|b^2|-2|a|\,|b|\cos t
}$$
Consider the function $f(t)$ on $[-\pi,\pi]$
$$\displaylines{f(t) =|a+b|^p+|a-b|^q\\ =\left (|a|^2+|b^2|+2|a|\,|b|\cos t\right )^{q/2}+\left (|a|^2+|b^2|-2|a|\,|b|\cos t\right)^{q/2}}$$
The function $f(t)$ is even, so we may restrict to $[0,\pi].$
We have
$$\displaylines{f'(t)=-q\,|a|\,|b|\,\sin t\,\left [\left (|a|^2+|b^2|+2|a|\,|b|\cos t\right )^{q/2-1}\\ - \left (|a|^2+|b^2|-2|a|\,|b|\cos t\right )^{q/2-1}\right ]}$$
Moreover $f'(t)\le 0$ for $0\le t\le \pi/2$ and $f'(t)\ge 0$ for $\pi/2\le t\le \pi$ (as $q>2$). So the maximum is attained at $0$ or $\pi.$
Next $$f(0)=f(\pi)=(|a|+|b|)^q+\big |\, |a|-|b|\,\big |^q,\quad |a|\ge |b|$$
Summarizing, we have obtained
$$|a+b|^q+|a-b|^q\le (|a|+|b|)^q+\big |\, |a|-|b|\,\big |^q $$ thus the problem has been reduced to real numbers.
A: Since the inequality in question is used primarily to derive bounds in $L_p$, namely Clarkson's inequality for $1<p<2$:
$$\Big\|\frac{f+g}{2}\Big\|^{p'}_p+\Big\|\frac{f-g}{2}\Big\|^{p'}_p\leq \Big(\frac12\big(\|f\|^p_p+\|g\|^p_p\Big)^{1/(p-1)},$$
I will present a short proof of the inequality in the OP for $\mathbb{C}$ based on  interpolation methods, namely the Riesz-Thorin interpolation theorem which I state now:

Theorem: Let $(X,\mathscr{F},\mu)$ and $(Y,\mathscr{G},\nu)$ be $\sigma$-finite measures spaces and $1\leq p_0,p_1\leq\infty$. Suppose $T:L_{p_0}(\mu)+L_{p_1}(\mu)\rightarrow L_{q_0}(\nu)+ L_{q_1}(\nu)$ is a linear operator such that
\begin{align}
\|Tf\|_{q_0}&\leq M_0\|f\|_{p_0},\qquad f\in L_{p_0}(\mu)\\
\|Tg\|_{q_1}&\leq M_1\|g\|_{q_1},\qquad g\in L_{p_1}(\mu)
\end{align}
For any $0<t<1$ define
$$\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1},\qquad \frac{1}{q_t}=\frac{1-t}{q_0}+\frac{t}{q_1}$$
Then, $T:L_{p_t}(\mu)\rightarrow L_{q_t}(\nu)$ is a bounded linear operator and
$$\|T\,h\|_{p_t}\leq M^{1-t}_0M^t_1\|h\|_{q_t}$$

Appealing to this theorem may appear as an overkill, however due to frat that the inequality of the OP is used primarily in the study of $L_p$ spaces, and that the interpolation theorem above is part of the curricula in integration theory, we might as well show a nice application of it. Also, elementary proof of the OP's inequality (using differentiation, or Calculus methods) are actually long and tricky, even in the real case (see for example, Hewitt, E. and Stronberg, K., Real and Abstract Analysis, Springer Verlag, GTM Vol 25, 1965, pp. 225-227). Finally,  versions of the Riesz-Thorin's theorem in the context of $\ell_p$ also appear in the classic book Hardy, G. et. al., Inequalities, Cambridge University, 1934, pp. 217-219.

Proof of the OP's inequality: Consider the operator $T$ on $\mathbb{C}^2$ defined as
$$T\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}a+b\\ a-b\end{pmatrix}
$$
Since all $p$-norms on Eucliean space are equivalent, we may ask ourselves what is the operator norm $\|T\|_{p,q}$ when we consider $T:(\mathbb{C},\|\,\|_p)\rightarrow(\mathbb{C},\|\;\|_q)$. Notice that for $p=2=q$
$$|a+b|^2+|a-b|^2=2(|a|+|b|^2)$$
ans so $\|T\|_{2,2}=\sqrt{2}$. On the other hand, for $p=1$ and $q=\infty$ we have $\max\{|a+b|,|a-b|\}\leq|a|+|b|$ with equality when for example $a=b=1$. Thus $\|T\|_{1,\infty}=1$. Thus, for $p_t,q_t$ with
$$\frac{1}{p_t}=\frac{1-t}{2}+t,\qquad \frac{1}{q_t}=\frac{1-t}{2}$$
$$\|T\|_{p_t,q_t}\leq 2^{(1-t)/2}$$
That is
\begin{align}
\Big(|a+b|^q+|a-b|^q\Big)^{1/q}\leq 2^{(1-t)/2}\big(|a|^p+|b|^p\big)^{1/p}\end{align}
Notice that $\frac{1}{2}<\frac{1}{p}<1$ and that $\frac1p+\frac1q=1$. Substituting $q=\frac{2}{1-t}$ yields the desired inequality
\begin{align}
|a+b|^q+|a-b|^q\leq 2 \big(|a|^p+|b|^p\big)^{q/p}=
 2 \big(|a|^p+|b|^p\big)^{q-1}\tag{1}\label{one}\end{align}

Comment: Clarkson's inequality
$$
|a+b|^q+|a-b|^q\leq 2^{q-1}(|a|^q+|b|^q),\qquad q\geq2$$
can be obtained by interpolating between $\|T\|_{2,2}$ and $\|T\|_{\infty,\infty}$; another approach is to use \eqref{one} and Hólder's inequality: Since $\frac1p+\frac1q=1$, $1<p<q$ and so,
$$
|a|^p+|b|^p=2\Big(\frac12 |a|^p\cdot1+\frac12 |b|^p\cdot1\Big)\leq 2\Big(\frac{1}{2}|a|^q+\frac{1}{2}|b|^q\Big)^{p/q}=2^{1-\tfrac{p}{q}}(|a|^q+|b|^q)^{p/q}
$$
Putting things together, we obtain
$$|a+b|^q+|a-b|^q\leq 2\big(|a|^p+|b|^p\big)^{q-1}\leq
2^{q/p}(|a|^q+|b|^q)=2^{q-1}(|a|^q+|b|^q)
$$

The idea of using interpolation to derive a simple proof of Clarkson's inequalities for $\mathbb{C}$ appears in the paper
Boas, R. P. Jr,, Some Uniformly Convex Spaces, Bull. Amer. Math. Soc. Vol.46 • No. 4 • April 1940., pp. 304-311
