Inequality for Olympiad students Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that
$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}} \ge 6$
My attempts you can see here: https://scontent-xsp1-2.xx.fbcdn.net/v/t1.0-9/90231854_2782055341915789_3356982430379540480_n.jpg?_nc_cat=101&_nc_sid=8024bb&_nc_ohc=pSSNCiYYb8kAX8lk9Cx&_nc_ht=scontent-xsp1-2.xx&oh=7f6cc512431574e8576250883ef88325&oe=5E9E41A9
 A: Another way (L.Hadassy, Y.Ilany).
If $a\geq b\geq c$ we have $$\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{c}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3b+\frac{1}{b}}$$ because it's $$(a-b)(b-c)\geq0.$$
Thus, 
$$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}.$$
If $a\geq c\geq b$ we have $$\sqrt{3b+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}\geq\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}$$ because it's $$(a-c)(c-b)\geq0,$$ which gives
$$\sum_{cyc}\sqrt{3a+\frac{1}{b}}\geq\sqrt{3a+\frac{1}{b}}+\sqrt{3b+\frac{1}{a}}+\sqrt{3c+\frac{1}{c}}.$$
Now we see that in any case we need to prove that:
$$\sqrt{3a+\frac{1}{c}}+\sqrt{3c+\frac{1}{a}}+\sqrt{3b+\frac{1}{b}}\geq6$$ or
$$\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)+\sqrt{3b+\frac{1}{b}}\geq6,$$
where $a\geq b\geq c$ or $c\geq b\geq a$.
Now, let $a+c=p=constant,$ $f(a,c)=\sqrt{3ac+1}\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}\right)$ and $F(a,c,\lambda)=f(a,c)+\lambda(a+c-p).$
Thus, in the minimum point we have $$\frac{\partial F}{\partial a}=\frac{\partial F}{\partial c}=0,$$ which gives
$$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial c}$$ or
$$(\sqrt{a}-\sqrt{c})(3\sqrt{a^3c^3}-\sqrt{ac}-a-c)=0.$$
1) $a=c$.
Thus, $b=3-2a,$ where $0<a<\frac{3}{2}$ and we need to prove that
$$2\sqrt{\frac{3a^2+1}{a}}+\sqrt{9-6a+\frac{1}{3-2a}}\geq6,$$ which is true.
2) $3\sqrt{a^3c^3}-\sqrt{ac}-a-c=0.$
Let $\sqrt{ac}=x$.
Thus, $a+c=3x^3-x,$ $b=3+x-3x^3$ and since $0<b\leq\frac{3}{2},$ we obtain: $0.932...<x\leq1.11...$ and we need to prove that:
$$\sqrt{3x^2+1}\cdot\sqrt{\frac{a+c+2\sqrt{ac}}{ac}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6$$ or
$$\frac{3x^2+1}{\sqrt{x}}+\sqrt{3(3+x-3x^3)+\frac{1}{3+x-3x^3}}\geq6,$$ which is true and it ends the proof.
