Verify the triangle inequality for $\lVert (u_1,u_2)\rVert := \sqrt[4]{u_1^4+u_2^4}$ In a lecture we were left to verify that the triangle inequality holds for 
$\lVert (u_1,u_2)\rVert := \sqrt[4]{u_1^4+u_2^4}$
In other examples I've been able to use the Cauchy Schwarz inequality to verify the triangle inequality, but I've tried expanding $\lVert (u_1+v_1,u_2+v_2)\rVert^4$ and it just becomes a mess. Could someone recommend a better approach? Cheers
 A: This is easier to write for $n $-tuples than it is for pairs. It's a beautiful story that depends on Hölders's inequality $$\sum_ja_jb_j\leq\left (\sum_ja_j^p\right ) ^{1/p}\left (\sum_jb_j^q\right ) ^{1/q} $$
for $a_j,b_j\geq0$ and $p,q>1$ with $\frac1p+\frac1q=1$ (proof at the end).
Then \begin{align}
\sum_j|u_j+v_j|^p&=\sum_j|u_j+v_j|\,|u_j+v_j|^{p-1}
\leq \sum_j|u_j|\,|u_j+v_j|^{p-1}+ \sum_j|v_j|\,|u_j+v_j|^{p-1} \\
&\leq\left (\sum_j|u_j|^p\right ) ^{1/p}\left (\sum_j|u_j+v_j|^{(p-1)q}\right ) ^{1/q}
+\left (\sum_j|v_j|^p\right ) ^{1/p}\left (\sum_j|u_j+v_j|^{(p-1)q}\right ) ^{1/q} \\
&= \left (\sum_j|u_j|^p\right ) ^{1/p}\left (\sum_j|u_j+v_j|^{p}\right ) ^{1/q}
+\left (\sum_j|v_j|^p\right ) ^{1/p}\left (\sum_j|u_j+v_j|^{p}\right ) ^{1/q} 
\end{align}
Now, since $1-1/q=1/p $,
$$
\left (\sum_j|u_j+v_j|^p\right)^{1/p}
=\frac { \left .\sum_j|u_j+v_j|^p\right.} { \left (\sum_j|u_j+v_j|^p\right)^{1/q}}\\
\leq
\left (\sum_j|u_j|^p\right ) ^{1/p} 
+
\left (\sum_j|v_j|^p\right ) ^{1/p}.
$$

To prove Hölders's inequality one needs Young's inequality. It all starts with the concavity of the log. For $t\in (0,1) $, $x,y\geq0$,
$$
t\log x+(1-t)\log y\leq \log(tx+(1-t)y).
$$
Exponentiating,
$$
x^ty^{(1-t)}\leq tx+(1-t)y.
$$
Now for $a,b\in\mathbb C $ and $p,q>1$ with $1/p+1/q=1$ we take $t=1/p $, $x=|a|^p $, $y=|b|^q $, and we get Young's inequality:
$$
|ab|\leq\frac1p\,|a|^p+\frac1q\,|b|^q.
$$
Finally, to Hölder's Inequality:
\begin{align}
\sum_j a_jb_j&=\left(\sum_ja_j^p\right)^{1/p}\left(\sum_jb_j^q\right)^{1/q}\,\sum_j\frac{a_j}{\left(\sum_ja_j^p\right)^{1/p}}\frac{b_j}{\left(\sum_jb_j^q\right)^{1/q}}\\
&\leq\left(\sum_ja_j^p\right)^{1/p}\left(\sum_jb_j^q\right)^{1/q}\,\sum_j\frac1p\,\frac{a_j^p}{\sum_ja_j^p}+\frac1q\,\frac{b_j^q}{\sum_jb_j^q}\\
&= \left(\sum_ja_j^p\right)^{1/p}\left(\sum_jb_j^q\right)^{1/q}\,\left(\frac1p+\frac1q\right)\\
&= \left(\sum_ja_j^p\right)^{1/p}\left(\sum_jb_j^q\right)^{1/q}.
\end{align}
