<s>[INST] $\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq \sqrt a+\sqrt b+ \sqrt c.$ I was practicing some inequality problems when I saw the following problem:
Problem: If $a,b,c$ are the three sides of a triangle, prove that $$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b}\leq \sqrt a+\sqrt b+ \sqrt c.$$
Solution: We assume, without loss of generality, that $a\geq b\geq c$ and notice that $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$. Then the rest of the solution is just Karamata.
But my question is on the statement: $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$. According to the definition of majorization, $\{a_i\}_{i=1}^n\succ \{b_i\}_{i=1}^n$ if and only if for all $1\leq k\leq n$, $\sum_{i=1}^ka_i\geq \sum_{i=1}^kb_i$ ( see here as reference). Of course, the inequality holds for $k=1$ and $k=3$ in our original case. But for $k=2$, we have $a+b-c+b+c-a=2b\leq a+b$. Then how come $(a+b-c,b+c-a,c+a-b)\succ (a,b,c)$? Is the solution above wrong or am I missing something?
 [/INST] For $a\geq b\geq c$ it should be $$(a+b-c,a+c-b,b+c-a)\succ(a,b,c).$$
Because $$a+b-c\geq a+c-b\geq b+c-a,$$ $$a+b-c\geq a,$$ $$a+b-c+a+c-b\geq a+b$$ and
$$a+b-c+a+c-b+b+c-a=a+b+c.$$
Now, we can use Karamata.
</s>