4. ĐK $x,y> 0$
$\frac{(a+b)^{2}}{c}+\frac{c^{2}}{a}\geq 4b$
Có $A=\frac{(a+b)^{2}}{c}+\frac{c^{2}}{a}\geq \frac{(a+b+c)^{2}}{c+a}$
mà $\frac{(a+b+c)^{2}}{c+a}\geq 4b\Leftrightarrow a^{2}+b^{2}+c^{2}-2bc-2ab+2ca\geq 0\Leftrightarrow (a-b+c)^{2}\geq 0$ (đúng)
Dấu ''='' $\Leftrightarrow \left\{\begin{matrix} a,b,c>0\\\frac{a+b}{c} =\frac{c}{a} \\ a+c=b \end{matrix}\right.\Leftrightarrow a=\frac{b}{3}=\frac{c}{2}$ (a>0)