Let $a=x+y$, $b=y+z$, $c=z+x$, then $a+b-c=2y$, $b+c-a=2z$, $c+a-b=2x$. Now the inequality becomes: \[ \sqrt{2x}+\sqrt{2y}+\sqrt{2z} \leq \sqrt{x+y}+\sqrt{y+z}+\sqrt{x+z} \] Since $\sqrt{\frac{(\sqrt{x})^2+(\sqrt{y})^2}{2}} \geq \frac{\sqrt{x}+\sqrt{y}}{2}$, so sum up and done. Equality holds if and only if $x=y=z$, that is $a=b=c$.

$(\sqrt{a+b-c} + \sqrt{b+c-a} )^2 = 2b + 2\sqrt{b^2 - (a-c)^2} \le 4b$ with equality when $a=c$.

Actually, $[a,b,c]$ is majorized by $[a+b-c, c+a-b,, b+c-a]$. So all kinds of triangle inequalities can be created using a convex/concave function. For example, let $f(x)=\frac{1}{x^2}$, then we have the inequality: \[ \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le \frac{1}{(a+b-c)^2}+\frac{1}{(c+a-b)^2}+\frac{1}{(b+c-a)^2}. \]

let $x=a+b-c,y=b+c-a.z=c+a-b$ then $\sqrt{a+b-c} + \sqrt{b+c-a}=\sqrt{x}+\sqrt{y} \leq\sqrt{2}\sqrt{x+y}=2\sqrt{b}$

Karamata only. Let $a\geq b\geq c$ $f(x)=\sqrt{x}$ is concave, and $[a,b,c]\succ [b+c-a,c+a-b,a+b-c]$, and we are done.

I had a doubt with Karamata in this problem. $f(x)=\sqrt{x}$ is a concave function. Let $a\geq b\geq c$, then $(a,b,c)\succ (b+c-a,c+a-b,a+b-c)$. Evan Chen's handout wrote: If $f$ is convex, and $(x_n)$ majorizes $(y_n)$ then \[f(x_1)+\dots + f(x_n) \geq f(y_1)+\dots f(y_n).\]The reverse inequality holds when $f$ is concave. Doesn't this mean $\sqrt{a}+\sqrt{b} + \sqrt{c} \leq \sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b}$? Could someone find the mistake in my argument.

$(a,b,c) \not \succ (a+b-c,c+a-b,b+c-a)$ (here $a \geq b \geq c \quad \text{so} \quad a+b-c \geq c+a-b \geq b+c-a$) as \(a \not \geq a+b-c\).

TheDarkPrince wrote: I had a doubt with Karamata in this problem. $f(x)=\sqrt{x}$ is a concave function. Let $a\geq b\geq c$, then $(a,b,c)\succ (b+c-a,c+a-b,a+b-c)$. Your second tuple is sorted incorrectly; we should have $a+b-c \ge c+a-b \ge b+c-a$ instead. Then $(a,b,c) \prec (a+b-c, c+a-b, b+c-a)$.

We also do Ravi's Substitution and obtain, that \[ \sqrt{x+y}+\sqrt{y+z}+\sqrt{x+z} \geq \sqrt{2x}+\sqrt{2y}+\sqrt{2z}. \]Squaring both sides and canceling terms, we obtain \[ \sqrt{(x+y)(x+z)}+\sqrt{(x+y)(y+z)}+\sqrt{(x+z)(y+z)} \geq 2\sqrt{xy}+2\sqrt{xz}+2\sqrt{yz}. \]By Cauchy-Schwarz, $$(x+y)(x+z)\geq (\sqrt{xy}+\sqrt{xz})^2\implies \sqrt{(x+y)(x+z)}\geq \sqrt{xy}+\sqrt{xz}.$$Similarly we do for others and adding them together we get desired inequality, equality occurs if and only if $x^2=yz$, $y^2=xz$ and $z^2=xz$, thus $x=y=z\implies a=b=c$. EDIT: 420th post, lets go, imm beast.

WLOG $a\ge b\ge c.$ Notice $(a,b,c)\succ (b+c-a,a+c-b,a+b-c)$ as $2a\ge b+c,$ $a+b\ge 2c,$ and $a+b+c\ge a+b+c.$ We are done by Karamata on $f(x)=\sqrt{x}.$ $\square$

Since $a,b,c$ are sides of a triangle, let $a=x+y,b=x+z,c=y+z$ for $x,y,z>0.$ Our inequality reduces to $$ \sum_{\text{cyc}}\sqrt{x+y}\geq \sum_{\text{cyc}}\sqrt{2x}$$By QM-AM, $$\sum_{\text{cyc}}\sqrt{\frac{x+y}{2}} \geq \sum_{\text{cyc}}\frac{\sqrt{x}+\sqrt{y}}{2}$$$$\sum_{\text{cyc}}\sqrt{x+y} \geq \sum_{\text{cyc}}\frac{\sqrt{2x}}{2}+\frac{\sqrt{2y}}{2},$$hence done.

Let $a=x+y,b=y+z,c=z+x.$ The inequality that we want to show becomes $$\sqrt{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\leq \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}.$$ Claim: For positive reals $x,y,$ we have $$\sqrt{x+y}\geq \frac{\sqrt{2}}{2}(\sqrt{x}+\sqrt{y}).$$Since this is homogenous, assert $x+y=2.$ Note that $\sqrt{x}$ is concave, so by Jensen's, we have $$\frac{\sqrt{2}}{2}(\sqrt{x}+\sqrt{y})\leq \frac{\sqrt{2}}{2}(2)=\sqrt{2}.$$ Summing the claim cyclically gives the desired conclusion. (equality when the triangle is equilateral)

question

Perform the substitution $a = y+z, b = z+x, c = x+y$. Then the inequality becomes $$\sum_{cyc} \sqrt{x+y} \ge \sum_{cyc} \sqrt{2x}.$$But from Jensen on $\sqrt{x}$, $$\sqrt{x+y} \ge \frac{\sqrt{2x} + \sqrt{2y}}{2}.$$Adding cyclically finishes. Equality holds iff $x = y = z \iff a = b = c$.