Lời giải:
a)
\(S=12(x^3+y^3)+16x^2y^2+34xy\)
\(=12[(x+y)^3-3xy(x+y)]+16x^2y^2+34xy\)
\(=12(1-3xy)+16x^2y^2+34xy=12+16x^2y^2-2xy\)
\(=(4xy-\frac{1}{4})^2+\frac{191}{16}\geq \frac{191}{16}\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} x+y=1\\ xy=\frac{1}{16}\end{matrix}\right.\Leftrightarrow (x,y)=(\frac{2+\sqrt{3}}{4}, \frac{2-\sqrt{3}}{4})\)
Vậy \(S_{\min}=\frac{191}{16}\) khi \(\Leftrightarrow (x,y)=(\frac{2+\sqrt{3}}{4}, \frac{2-\sqrt{3}}{4})\) và có hoán vị.
b)
\(A=5(x^3+y^3)+12xy+4x^2y^2\)
\(=5[(x+y)^3-3xy(x+y)]+12xy+4x^2y^2\)
\(=5(1-3xy)+12xy+4x^2y^2\)
\(=5+4x^2y^2-3xy\)
Áp dụng BĐT Cô-si: $1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$A=4x^2y^2-3xy+5=xy(4xy-1)-\frac{1}{2}(4xy-1)+4,5=(xy-\frac{1}{2})(4xy-1)+4,5$
Vì $xy\leq \frac{1}{4}\Rightarrow 4xy-1\leq 0; xy-\frac{1}{2}< 0\Rightarrow (xy-\frac{1}{2})(4xy-1)\geq 0$
$\Rightarrow A=(xy-\frac{1}{2})(4xy-1)+4,5\geq 4,5$
Vậy $A_{\min}=4,5$ khi $x=y=\frac{1}{2}$