How to prove a "generalization" of the "Clarkson inequality"? So let $s\leq r$ and $s\leq2$, and $x,y\in\mathbb{R}$. I am interested in how to prove the following inequality:
\begin{equation}
\left( \lvert x+y \rvert^r + \lvert x-y \rvert^r \right)^{1/r} \leq 2^{1-1/s}\left( \lvert x \rvert^s + \lvert y \rvert^s \right)^{1/s}.
\end{equation}
In fact, the original problem is to prove that for required $r$ and $s$ ($s\leq p\leq r$ and $s\leq2$), every $f$ and $g$ in $L_p(X)$ satisfy the inequality above. I realized that since every normed value is real, the inequality for $\mathbb{R}$ may be a reasonable reduction, but then I am stuck on proving it. Thank you and I appreciate any insights!
 A: The left-hand side is increasing in $r$, see for example How do you show monotonicity of the $\ell^p$ norms?. Therefore it suffices to prove the inequality for the smallest possible value of $r$, that is for $r=s$:
$$
\left( \lvert x+y \rvert^s + \lvert x-y \rvert^s \right)^{1/s} \leq 2^{1-1/s}\left( \lvert x \rvert^s + \lvert y \rvert^s \right)^{1/s}.
$$
This Clarkson type inequality holds for $s \ge 2$, see for example:

*

*Showing $\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$

*Three related inequalities (the first being $2(|a|^p + |b|^p) \leq |a + b|^p + |a - b|^p \leq 2^{p-1}(|a|^p + |b|^p)$)
On the other hand, the inequality does not hold in general for $0 < s < 2$, as can be seen by setting $x=1$ and $y=0$.
A: For $1< p\leq 2$  we have the inequalities
$$\begin{align}
\Big(|a+b|^{p'}+|a-b|^{p'}\Big)^{1/{p'}}\leq 2^{1-\tfrac1p}\Big(|a|^p+|b|^p\Big)^{1/p}\leq2^{1/p}\Big(|a|^{p'}+|b|^{p'}\Big)^{1/{p'}}\tag{0}\label{clark}\end{align}$$
where $\frac{1}{p}+\frac{1}{p'}=1$. These are known as Clarkson's inequality (see comments below).
For the question in the OP, consider the case $s\leq 2$ and $s'\leq r$. The decreasing monotonicity of $p$-norms along with inequality \eqref{clark} implies
$$\Big(|a+b|^r+|a-b|^r\Big)^{1/r}\leq\Big(|a+b|^{s'}+|a-b|^{s'}\Big)^{1/{s'}}\leq 2^{1-\tfrac1s}\Big(|a|^s+|b|^s\Big)^{1/s}$$
which shows that the statement in the OP holds for $s\leq 2$ and $s'\leq r$. Other than that, the statement in the OP may fail.

Comments:
Inequality \eqref{clark} can be obtained by interpolation (Riesz-Thorin) by considering the linear operator $T:(\mathbb{C}^2,\|\;\|_p)\rightarrow(\mathbb{C}^2,\|\;\|_q)$ defined
as
$$
T\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}a+b\\ a-b\end{pmatrix}$$
Since all $p$ norms, $p\geq1$, are equivalent, $T$ is a bounded operator, that is
$$\|T\|_{p,q}=\sup_{\|v\|_p\neq0}\|Tv\|_q<\infty$$
It is easy to check that $\|T\|_{2,2}=\sqrt{2}$, $\|T\|_{1,0}=1$, and $\|T\|_{\infty,\infty}=2=\|T\|_{1,1}$. Using interpolation between $\|T\|_{2,2}$ and $\|T\|_{1,0}$ yields \eqref{clark}.
Another useful inequalities are obtain by interpolating between $\|T\|_{2,2}$ and $\|T\|_{\infty,\infty}$ which yields
$$\begin{align}
\Big(|a+b|^r+|a-b|^r\Big)^{1/r}\leq 2^{1-\tfrac1r}\big(|a|^r+|b|^r\big)^{1/r}\leq 2^{1-\tfrac1r}\big(|a|^s+|b|^s\big)^{1/s}\tag{1}\label{c\lark1}
\end{align}$$
for all $1\leq s\leq 2\leq  r$, and by interpolating between $\|T\|_{1,1}$ and $\|T\|_{2,2}$ which yields
$$\begin{align}
\Big(|a+b|^r+|a-b|^r\Big)^{1/r}\leq 2^{1/r}\big(|a|^r+|b|^r\big)^{1/r}\leq 2^{1/r}\big(|a|^s+|b|^s\big)^{1/s}\tag{2}\label{clark2}
\end{align}$$
for $1<s<r\leq2$.

Further generalizations cab be seen in a the paper Maligranda, L and Persson, L. E., On Clarkson's inequalities and Interpolation, Math Nachr. 166 (1992) pp. 187-197.
