RMO Question

Applying AM-GM to $a+b$ , $b+c$ and $c+a$ and $\frac{1}{a+b}$ , $\frac{1}{b+c}$ and $\frac{1}{c+a}$ ,

$\frac23 \left( a+b+c \right) \ge \sqrt [ 3 ]{ \left( a+b \right) \left( b+c \right) \left( c+a \right) }$

$\frac { 1 }{ 3 } \left( \frac { 1 }{ a+b } +\frac { 1 }{ b+c } +\frac { 1 }{ c+a } \right) \ge \sqrt [ 3 ]{ \frac { 1 }{ \left( a+b \right) \left( b+c \right) \left( c+a \right) } }$

On multiplying these inequalities,

$\begin{matrix} \frac { 2 }{ 9 } \left( a+b+c \right) \left( \frac { 1 }{ a+b } +\frac { 1 }{ b+c } +\frac { 1 }{ c+a } \right) & \ge & 1 \\ \frac { a+b+c }{ a+b } +\frac { a+b+c }{ b+c } +\frac { a+b+c }{ c+a } & \ge & \frac { 9 }{ 2 } \\ \frac { c }{ a+b } +1+\frac { a }{ b+c } +1+\frac { b }{ c+a } & \ge & \frac { 9 }{ 2 } \\ \frac { a }{ b+c } +\frac { b }{ c+a } +\frac { c }{ a+b } & \ge & \frac { 3 }{ 2 } \end{matrix}$

Since $a^2+1 \ge 2a$ , $\frac { { a }^{ 2 }+1 }{ b+c } \ge \frac { 2a }{ b+c }$ . Similarly, $\frac { { b }^{ 2 }+1 }{ c+a } \ge \frac { 2b }{ c+a }$ and $\frac { { c }^{ 2 }+1 }{ a+b } \ge \frac { 2c }{ a+b }$ . On adding them,

\frac { { a }^{ 2 }+1 }{ b+c } +\frac { { b }^{ 2 }+1 }{ c+a } +\frac { { c }^{ 2 }+1 }{ a+b } \ge \frac { 2a }{ b+c } +\frac{2b}{ c+a } +\frac { 2c }{ a+b }

As

$\frac { a }{ b+c } +\frac { b }{ c+a } +\frac { c }{ a+b } \ge \frac { 3 }{ 2 } \\ 2\left( \frac { a }{ b+c } +\frac { b }{ c+a } +\frac { c }{ a+b } \right) \ge 3$

Therefore by transitive property,

$\frac { { a }^{ 2 }+1 }{ b+c } +\frac { { b }^{ 2 }+1 }{ c+a } +\frac { { c }^{ 2 }+1 }{ a+b } \ge 3$

Also, when $a =b=c=1$ , equality occurs, proving that 3 is indeed the minimum value.