<s>[INST] A problem from KVS 2014 I was doing the the following problem-
Prove that $$\frac { \sqrt { a+b+c } +\sqrt { a }  }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b }  }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c }  }{ a+b } \ge \frac { 9+3\sqrt { 3 }  }{ 2\sqrt { a+b+c } }. $$
I normalized this to $a+b+c=1$ and simplified to get-
$$\frac { 1 }{ 1-\sqrt { a }  } +\frac { 1 }{ 1-\sqrt { b }  } +\frac { 1 }{ 1-\sqrt { c }  } \ge \frac { 9+3\sqrt { 3 }  }{ 2 }.$$
By Titu's lemma,
$$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a }  } +\frac { 1 }{ 1-\sqrt { b }  } +\frac { 1 }{ 1-\sqrt { c }  } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } .$$
However the RHS is maximized when  $\sqrt { a } +\sqrt { b } +\sqrt { c } $ is maximized which is at $\sqrt { 3 } $,
$$\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a }  } +\frac { 1 }{ 1-\sqrt { b }  } +\frac { 1 }{ 1-\sqrt { c }  } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \le \frac { 9 }{ 3-\sqrt { 3 }  } =\frac { 9+3\sqrt { 3 }  }{ 2 } .$$
What went wrong?
You can view the rest of the problems here.
 [/INST] we consider $f(x)=\frac{1}{1-\sqrt{x}}$ for $0<x<1$ and the tangent line at the point $x=\frac{1}{3}$ we get $y=\frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$. We have for $0<x<1$ $$f(x)\geq \frac{3}{4}\sqrt{3}(2+\sqrt{3})x+\frac{3}{4}$$. Thus we get $$\frac{1}{1-\sqrt{a}}+\frac{1}{1-\sqrt{b}}+\frac{1}{1-\sqrt{c}}\geq \frac{3}{4}\sqrt{3}(2+\sqrt{3})(a+b+c)+\frac{9}{4}=\frac{1}{2}(9+3\sqrt{3})$$
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