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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?

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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 $$\sqrt{c^2+e^2}+\sqrt{d^2+f^2}=\sqrt{c^2+e^2+d^2+f^2+2\sqrt{(c^2+e^2)(d^2+f^2)}}\geq$$ $$\geq\sqrt{c^2+e^2+d^2+f^2+2(cd+ef)}=\sqrt{(c+d)^2+(e+f)^2}\geq\sqrt{a^2+b^2}$$

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    Sorry if this is a little off topic here, but I see you are very good at solving inequalities and I would like to ask you something. I am 2-3 year math undergraduate student. Do you have any fun books to reccomend on inequalities for somebody with zero experience in inequalities like me? – 2017-02-25
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    @Ovi I think the best book to learn inequalities is the following forum:https://www.artofproblemsolving.com/community/c6t243f6_inequalities Just read it! – 2017-02-25
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    I've tried to show that $D(x,y)=d_1(x,y)+d_2(x,y)$ with both $d_1$ and $d_2$ metrics, are also a metric – 2017-02-25
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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
$$\|X\| \le \|Y_1+Y_2\| \le \|Y_1\|+\|Y_2\|$$

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    You are using that $(\mathbb{R}^2,||•||)$ are a metric space right? – 2017-02-24
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    @Ragnar1204, in fact $(\mathbb{R}^2,||•||)$ – 2017-02-24
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    Yes, my fault... – 2017-02-24
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    Sorry by my question. But... There's an order in the euclidean plane? Cuz I can't follow the inequality between vectors... Right? – 2017-02-25
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    @Ragnar1204, the inequality is componentwise. It means that each coordinate of one vector is less than of the other. – 2017-02-25
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    @AndreiKulunchakov Maybe you should make that more clear in the answer, since it's not too common a notation ($\le$ is generally used for total order relations, while component-wise inequality is only a partial order). It should also be noted that the conclusion only follows when $0 \le X$. – 2017-02-25