Solve for $x$

Let the Left hand side of the inequality be called P . P can be written as:

$\frac { { a }^{ 2 } }{ a(2c+b) } +\frac { { b }^{ 2 } }{ b(2a+c) } +\frac { { c }^{ 2 } }{ c(2b+a) }$

By Titu's Lemma

$P\ge \frac { (a+b+c)^{ 2 } }{ 3(ab+bc+ca) } =1\quad ...**$

$**..\quad \frac { 1 }{ 2 } \left[ { (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (c-a) }^{ 2 } \right] \ge 0$

This is true by Trivial Inequality

$\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca\ge 0\Rightarrow { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+2ab+2bc+2ca\ge 3ab+3bc+3ca$

$\Rightarrow { (a+b+c) }^{ 2 }\ge 3(ab+bc+ca)\Rightarrow \frac { { (a+b+c) }^{ 2 } }{ 3(ab+bc+ca) } \ge 1$

Hence $x=\boxed { 1 }$ with equality when $a=b=c$