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I want to show the following elementary inequality:

$$((|a|+|b|)^p + (|c|+|d|)^p)^{\frac{1}{p}} \le (|a|^p+|c|^p)^{\frac{1}{p}} + (|b|^p+|d|^p)^{\frac{1}{p}}$$ and we have $p \ge 1$. Does anybody have an idea? It would be pretty useful if anybody could say something about this inequality.

If this one is easier for you, then it would also be sufficient to prove this one: $$((|a+b|)^p + (|c+d|)^p)^{\frac{1}{p}} \le (|a|^p+|c|^p)^{\frac{1}{p}} + (|b|^p+|d|^p)^{\frac{1}{p}}$$

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Your inequality follows directly from the Minkowski inequality (proof). $$ (|a+b|^p+|c+d|^p)^{1/p}=\left\|\begin{pmatrix} a+b\\c+d \end{pmatrix}\right\|_p\leq \left\|\begin{pmatrix} a\\c \end{pmatrix}\right\|_p+\left\|\begin{pmatrix} b\\d \end{pmatrix}\right\|_p=(|a|^p+|c|^p)^{1/p}+(|b|^p+|d|^p)^{1/p} $$

If a,b,c,d were positive u have exactly your case.

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