I don't know if there is a nicer solution if $a,b,c$ are sides of triangle, but from the following proof we can see that inequality holds for all nonnegative reals $a,b,c$ such that $a\ge b\ge c$. First, we define $a=x^2, \; b=y^2, \; c=z^2$. Let's fix now $x$ and $z$ and we will prove that $f(y)=x\sqrt{x^2-xy+y^2}+y\sqrt{x^2-xz+z^2}+z\sqrt{y^2-yz+z^2}-x^2-y^2- z^2$ is concave function. Our starting inequality is equivalent to: $f(y)\ge 0$ $f''(y)= \frac{3x^3}{4(x^2-xy+y^2)^{\frac{3}{2}}}+ \frac{3z^3}{4(z^2-zy+y^2)^{\frac{3}{2}}}-2$ From condiiton $x\ge y\ge z$ we have: $x^2-xy+y^2 \ge \frac{3}{4}x^2$ and $y^2-yz+z^2\ge z^2$, so: $f''(y) \leq \sqrt{\frac{4}{3}}+\frac{3}{4}-2<0$ Hence $f(y)$ reaches minimum when $x=y$ or when $y=z$. Since inequality remains unchanged when we swap $x$ and $z$, it's enough to prove inequality when $x=y$.Inequality is in this case equivalent to: $\sqrt{x^2-xz+z^2}(x+z)\ge x^2+z^2$ $\Leftrightarrow \sqrt{(x^3+z^3)(x+z)}\ge x^2+z^2$. What is cauchy-schwarz inequality. Therefore, inequality is proved. Equality holds iff $x=y=z$ or $x=y, \; z=0$ or $y=z=0$ with all permutations. $\blacksquare$

I forgot about this inequality after I posted my first proof because of some obligations and forgot to post more beautiful proof for actual problem, when $a,b,c$ are sides of a triangle. Proof: It's easy to prove that, with given conditions, arrays: $(a,b,c)$ and $(a-\sqrt{ab}+b, \; a-\sqrt{ac}+c, \; b-\sqrt{bc}+c)$ are similarly sorted. For example: $a-\sqrt{ab}+b \ge a-\sqrt{ac}+c \Longleftrightarrow (\sqrt{b}-\sqrt{c})\cdot \frac{b+c+2\sqrt{bc}-a}{\sqrt{b}+\sqrt{c}+\sqrt{a}} \ge 0$ what is obviously true and we can prove in the same way second inequality. Therefore, from rearrangement and cauchy-schwarz inequality, we have: $2 \left( \sqrt{a(a-\sqrt{ab}+b)}+\sqrt{b(a-\sqrt{ac}+c)}+\sqrt{c(b-\sqrt{bc}+c)} \right) \\ \ge \sum (\sqrt{a}+\sqrt{b})\sqrt{a-\sqrt{ab}+b}= \sum \sqrt{(\sqrt{a}+\sqrt{b})(a^{\frac{3}{2}}+b^{\frac{3}{2}})} \ge \sum (a+b) \\=2(a+b+c)$. Hence, inequality is proved. Equality holds when $a=b=c$. $\blacksquare$

duanKHTN wrote: you made 1 big mistake Sorry I stupid english And can you tell me where I made it? -.-

Dear MathUniverse, Your solution is very NICE! Congratulations !

mlm95 wrote: Let $a,b,c$ be sides of a triangle such that $a\geq b \geq c$. prove that: $\sqrt{a(a+b-\sqrt{ab})}+\sqrt{b(a+c-\sqrt{ac})}+\sqrt{c(b+c-\sqrt{bc})}\geq a+b+c$ This inequality holds for any positive real numbers $a,\,b,\,c$ satisfying $a \ge b \ge c.$ Indeed, setting $x=\sqrt{a},\,y=\sqrt{b}$ and $z=\sqrt{c},$ we can rewrite the inequality as \[x\sqrt{x^2-xy+y^2}+y\sqrt{x^2-xz+z^2}+z\sqrt{y^2-yz+y^2} \ge x^2+y^2+z^2.\] For any $u,\,v >0,$ by the Cauchy-Schwarz inequality, we have \[\sqrt{u^2-uv+v^2}=\frac{\sqrt{(u^3+v^3)(u+v)}}{u+v} \ge \frac{u^2+v^2}{u+v}.\] Using this inequality, it suffices to prove that \[\frac{x(x^2+y^2)}{x+y}+\frac{y(z^2+x^2)}{z+x}+\frac{z(y^2+z^2)}{y+z} \ge x^2+y^2+z^2.\] This inequality is equivalent to \[\frac{xy(y-x)}{x+y}+\frac{y\left[ z(z-y)+x(x-y)\right]{z+x}+\frac{yz(y-z)}{y+z} \ge 0,\] or \[xy(x-y) \left(\frac{1}{x+z}-\frac{1}{x+y}\right)+yz(y-z)\left(\frac{1}{y+z}-\frac{1}{x+z}\right) \ge 0,\] which be rewritten as \[\frac{xy(x-y)(y-z)}{(x+y)(x+z)}+\frac{yz(x-y)(y-z)}{(x+z)(y+z)} \ge 0,\] which is obviously true.