An algebra problem by Pishi Meow

First by AM-GM $\left( \dfrac { a+b }{ 2c } \right) ^{ 2 }+\left( \dfrac { c+b }{ 2a } \right) ^{ 2 }+\left( \dfrac { a+c }{ 2b } \right) ^{ 2 }\ge 3\quad \cdot \sqrt [ 3 ]{ \left( \dfrac { (a+b)(b+c)(c+a) }{ 8abc } \right) ^{ 2 } }$

now we will prove that $(a+b)(b+c)(c+a)\geq 8abc$

This is equivalent to $\dfrac { a+b }{ \sqrt { ab } } \times \dfrac { c+b }{ \sqrt { cb } } \times \dfrac { a+c }{ \sqrt { ac } } \geq 2.2.2$

By AM-GM the inequality holds only if $a=b=c \therefore (a+b)(b+c)(c+a)\geq 8abc$

$\Rightarrow \left( \dfrac { a+b }{ 2c } \right) ^{ 2 }+\left( \dfrac { c+b }{ 2a } \right) ^{ 2 }+\left( \dfrac { a+c }{ 2b } \right) ^{ 2 }\ge 3\quad \cdot \sqrt [ 3 ]{ \left( \dfrac { 8abc }{ 8abc } \right) ^{ 2 } } \\ \Rightarrow \left( \dfrac { a+b }{ 2c } \right) ^{ 2 }+\left( \dfrac { c+b }{ 2a } \right) ^{ 2 }+\left( \dfrac { a+c }{ 2b } \right) ^{ 2 }\ge 3\times 1$

So minimum value of $\left( \dfrac { a+b }{ c } \right) ^{ 2 }+\left( \dfrac { c+b }{ a } \right) ^{ 2 }+\left( \dfrac { a+c }{ b } \right) ^{ 2 }$ will be $3\times4$

$\therefore$ the answer is $\huge\boxed{\color{royalblue}{12}}$

Call the expression P, by Cauchy-Schwarz and AM-GM we have $3A\geq \bigg(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\bigg)^2\geq 6^2$ $\Leftrightarrow A\geq 12$ The equality holds when $a=b=c$

using 3 variable AM-GM inequality you can observe that

$\sum\limits_{cyc} \left(\frac{a+b}{2c}\right)^2 \ge 3\sqrt[3]{\left(\frac{(a+b)(b+c)(c+a)}{8abc}\right)^2} \ge 3$

where $\sum_{cyc}$ denotes sums over cyclic permutations of the symbols $a,b,c$ . also i used the well known inequality which states

$(a+b)(b+c)(c+a)\ge 8abc$

which can be proved using AM-GM inequality. so we proved minimum value of $\sum\limits_{cyc} \left(\frac{a+b}{2c}\right)^2$ is 3 .So minimum value of $\sum\limits_{cyc} \left(\frac{a+b}{c}\right)^2$ will be 12.