gt$\Leftrightarrow$$2\left[ {(a+b-\frac{5}{2})^{2}+(c-\frac{5}{2})^{2}} \right] =25$
Áp dụng BĐT Bunhiacopxki:$25\geq (a+b+c-5)^{2}$$\Rightarrow$$a+b+c\leq10$
Ta có:$A=a+b+c+48(\frac{12}{2.\sqrt{12}.\sqrt{a+10}}+\frac{12}{3.2.2.\sqrt[3]{b+c}})$
Áp dụng BĐT Cauchy:$2.\sqrt{12}.\sqrt{a+10} \leq a+22$
$3.2.2.\sqrt[3]{b+c}\leq b+c+16$
$\Rightarrow$$\frac{12}{2.\sqrt{12}.\sqrt{a+10}}+\frac{12}{3.2.2.\sqrt[3]{b+c}}\geq \frac{12}{a+22}+\frac{12}{b+c+16}\geq \frac{48}{a+b+c+38}$
$\Rightarrow$$A\geq a+b+c+\frac{48^{2}}{a+b+c+38}$
Đặt $t=a+b+c$,$0<t\leq10$
$\Rightarrow$$A\geq t+\frac{48^{2}}{t+38}=f(t)$
BBT$\Rightarrow$$A\geq58$
Dấu''='' xra$\Leftrightarrow$$a=2;b=3;c=5$