How to show in $R^2$, the $f(x,y)=(x^4+y^4)^{1/4}$ defines a norm on $R^2$. How to show in $R^2$, the $f(x,y)=(x^4+y^4)^{1/4}$ defines a norm on $R^2$, i.e., how to prove the inequality:
$$((x_1+x_2)^4+(y_1+y_2)^4)^{1/4}\le (x_1^4+y_1^4)^{1/4}+(x_2^4+y_2^4)^{1/4}$$

There is a nice proof of this Minkowski by Holder:
Let $x_1=a$, $y_1=b$, $x_2=c$ and $y_2=d$.
Thus, we need to prove that
$$\sqrt[4]{a^4+b^4}+\sqrt[4]{c^4+d^4}\geq\sqrt[4]{(a+b)^4+(c+d)^4}.$$
Indeed, by Holder we obtain:
$$\left(\sqrt[4]{a^4+b^4}+\sqrt[4]{c^4+d^4}\right)^4=$$
$$=a^4+b^4+4\sqrt[4]{(a^4+b^4)^3(c^4+d^4)}+6\sqrt{(a^4+b^4)(c^4+d^4)}+$$
$$+4\sqrt[4]{(a^4+b^4)(c^4+d^4)^3}+c^4+d^4\geq$$
$$\geq a^4+b^4+4(a^3c+b^3d)+6(a^2c^2+b^2d^2)+$$
$$+4(ac^3+bd^3)+c^4+d^4=(a+c)^4+(b+d)^4.$$
