An inequality: from the complex to the real case. Let $p\in(1,2]$ and $q\in[2,\infty)$ be its conjugate exponent, then for $z,w\in\mathbb{C}$ the following inequality holds
$$
\Large \left|\frac{z+w}{2}\right|^q+\left|\frac{z-w}{2}\right|^q\leq\left[\frac{1}{2}\left(|z|^p+|w|^p\right)\right]^{\frac{1}{p-1}}\tag 1.
$$
The above inequality is a direct conseguence of the following inequality

If $1<p\le 2$ and $0\le t\le 1$, then $$\Large\left(\frac{1+t}{2}\right)^q+\left(\frac{1-t}{2}\right)^q\le \left(\frac{1}{2}+\frac{1}{2}t^p \right)^{\frac{1}{p-1}}\tag 2,$$ where $q=p/(p-1)$ is the exponent conjugate to $p$.

Question How can I get $(1)$ in the real case using the inequality $(2)$?
 A: $|\frac{z+w}{2}|^q+|\frac{z-w}{2}|^q =|z|^{q} \frac {(1+t)^{q}+(1-t)^{q}} {2^{q}}$ where $t=|\frac  w z|$. [Note that $t=\frac  w z$ or  $t=-\frac  w z$]. Hence,  $|\frac{z+w}{2}|^q+|\frac{z-w}{2}|^q \leq |z|^{q}(\frac 1  2+\frac 1 2 t^{p})^{2/(p-1)}$. Just put $t=|\frac  w z|$ and bring $|z|^{q}$ inside.
A: Remark: I think you want to prove (1) for complex numbers using (2).
It is easy to prove that
$$(|z| + |w|)^2 + (|z| - |w|)^2
= |z + w|^2 + |z - w|^2,$$
$$(|z| + |w|)^2 \ge |z \pm w|^2.$$
Let $f(x) = x^{q/2}$.
Since $q = p/(p - 1) \ge 2$,
$f(x)$ is convex on $x\ge 0$. Using Karamata's inequality, we have
$$f\Big((|z| + |w|)^2\Big)
+ f\Big((|z| - |w|)^2\Big)
\ge f(|z + w|^2) + f(|z - w|^2)$$
that is
$$(|z| + |w|)^q + \Big||z| - |w|\Big|^q
\ge |z + w|^q + |z - w|^q.$$
Thus, it suffices to prove that
$$\left(\frac{|z| + |w|}{2}\right)^q + \left|\frac{|z| - |w|}{2}\right|^q \le \left(\frac{|z|^p + |w|^p}{2}\right)^{1/(p - 1)}.$$
WLOG, assume that $|z| \ge |w|$ and $|z| > 0$. Then $|w|/|z| \in [0, 1]$.
It suffices to prove that
$$\left(\frac{1 + |w|/|z|}{2}\right)^q + \left|\frac{1 - |w|/|z|}{2}\right|^q \le \left(\frac{1 + |w|^p/|z|^p}{2}\right)^{1/(p - 1)}$$
which is true using (2).
