Let $a,b,c>0$ and $a+b+c=9$. Prove that: $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\geq ab+bc+ca.$$ (Edit) I'm sorry. The problem should have beeen: Let $a,b,c\geq1$ and $a+b+c=9$. Prove that: $$(\sqrt{a}+\sqrt{b}+\sqrt{c})^2\geq ab+bc+ca.$$
Since $(a+b+c)2\geq3(ab+bc+ca)$ or $ab+bc+ca\leq27$, all we need to do now is to prove $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\geq9$, which I cannot.
Let $a=2x+1$, $b=2y+1$ and $c=2z+1$.
Thus, $x$, $y$ and $z$ are non-negatives, $x+y+z=3$ and we need to prove that: $$\sum_{cyc}\sqrt{2x+1}\geq\sqrt{\sum_{cyc}(2x+1)(2y+1)}$$ or $$\sum_{cyc}\left(2x+1+2\sqrt{(2x+1)(2y+1)}\right)\geq\sum_{cyc}(4xy+4x+1)$$ or $$\sum_{cyc}(4x-4xy)\geq\sum_{cyc}\left(2x+1-2\sqrt{(2x+1)(2y+1)}+2y+1\right)$$ or $$4\sum_{cyc}(x^2+2xy-3xy)\geq3\sum_{cyc}\left(\sqrt{2x+1}-\sqrt{2y+1}\right)^2$$ or $$\sum_{cyc}(x-y)^2\left(1-\frac{6}{\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2}\right)\geq0.$$ Now, let $x\geq y\geq z$.
Thus, $$y^2\sum_{cyc}(x-y)^2\left(1-\frac{6}{\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2}\right)=$$ $$=y^2\sum_{cyc}(x-y)^2\left(1-\frac{6}{2x+2y+2+2\sqrt{(2x+1)(2y+1)}}\right)\geq$$ $$\geq y^2\sum_{cyc}(x-y)^2\left(1-\frac{3}{x+y+2}\right)=y^2\sum_{cyc}\frac{(x-y)^2(x+y-1)}{x+y+2}=$$ $$=y^2\sum_{cyc}\frac{(x-y)^2(3x+3y-x-y-z)}{3x+3y+2(x+y+z)}\geq y^2\sum_{cyc}\frac{(x-y)^2(x+y-z)}{5x+5y+2z}\geq$$ $$\geq\frac{y^2(x-z)^2(x+z-y)}{5x+5z+2y}+\frac{y^2(y-z)^2(y+z-x)}{5y+5z+2x}\geq$$ $$\geq\frac{x^2(y-z)^2(x-y)}{5x+5z+2y}+\frac{y^2(y-z)^2(y-x)}{5y+5z+2x}=$$ $$=\frac{(x-y)^2(y-z)^2(2x^2+2y^2+7xy+5xz+5yz)}{(5x+5z+2y)(5y+5z+2x)}\geq0$$ and we are done!