<s>[INST] Checking if a functional $F(x)$ is a norm in $\mathbb{R}^2$ There is a functional given:
$$F(x) = \sqrt{2x_1^2 + 3x_2^2}$$
Of course $x\in\mathbb{R}^2 \rightarrow x = (x_1, x_2)$
It is easy to check that:
1) $\forall x \in \mathbb{R}^2$ $F(x) \ge 0$
2) $F(x) = 0 \iff x = 0$
3) $ F(\lambda x) = | \lambda | F(x)$
I failed however to show that $F(x+y) \le F(x) + F(y)$. How should it be proven?
 [/INST] You can just use the Cauchy-Schwartz inequality:
\begin{align}
F(x+y)^2 &= 2(x_1+y_1)^2+3(x_2+y_2)^2 \\
&= \underbrace{2x_1^2 + 3x_2^2}_{F(x)^2} + \underbrace{2y_1^2+3y_2^2}_{F(x)^2}+2(2x_1y_1 + 3x_2y_2)\\
&\stackrel{CSB}{\le} F(x)^2 + F(y)^2 + 2\sqrt{2x_1^2+3x_2^2}\sqrt{2y_2^2+3y_2^2}\\
&= F(x)^2 + F(y)^2 + 2F(x)F(y)\\
&= (F(x) + F(y))^2
\end{align}
</s>