Triangular inequality extended Let $a,b,c,d,e,f\in\mathbb{R}$ all positive or zero, such that $a\leq c+d$ and $b\leq e+f$ show that:
$$\sqrt{a^2+b^2}\leq\sqrt{c^2+e^2}+\sqrt{d^2+f^2}$$
Some hint or auxiliar inequality that would help? I've done many attempts but write them will take a lot of time, I tried to use the artihmetic and geometric mean, minkowski but no one works (or fits) can you help me?
 A: Just $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}\geq\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$
If I don't see the triangle inequality I can make the following.
By C-S
$$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}=\sqrt{c^2+e^2+d^2+f^2+2\sqrt{(c^2+e^2)(d^2+f^2)}}\geq$$
$$\geq\sqrt{c^2+e^2+d^2+f^2+2(cd+ef)}=\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$
A: Denote vectors $X = (a,b), Y_1 = (c,e), Y_2 = (d,f)$. Then we have $X \le Y_1+Y_2$. It follows with $\|X\| \le \|Y_1+Y_2\| \le \|Y_1\|+\|Y_2\|$
Edit
Thanks to the comment from dxiv. Of course, it should be noted that the components of $X$ and $Y_i$ are positive. Otherwise we do not have the implication $\|X\| \le \|Y_1+Y_2\|$ from $X \le Y_1+Y_2$. For example, it is false for $X = (-2,-2), Y_1  = (1,1)~\textrm{and}~Y_2 = (-1,-1)$.
To be more precise on my proof, $X \le Y_1+Y_2$ means only componentwise order on vectors (i.e. partial order). So that each component of $X$ is smaller than the corresponding component of $Y_1+Y_2$. 
Why do we have $\|X\| \le \|Y_1+Y_2\|$ in this case? It could be verified by hands from
$$\sqrt{x_1^2+x_2^2} \le \sqrt{y_1^2+y_2^2},$$
for $y_i = \left(Y_1+Y_2\right)_i$ as an implication from componentwise order. It remains then to apply triangle inequality for the Euclidean norm and get
$$\|X\| \le \|Y_1+Y_2\| \le \|Y_1\|+\|Y_2\|$$
