AM-GM:\(x^2+4\ge4x\);\(y^2+4\ge4y\)
\(\Rightarrow VT\ge\left(4x+4y+4\right)\left(4x+4y+4\right)=\left(4x+4y+4\right)^2\)
Ta có:\(\left(3x+5y+4\right)\left(5x+3y+4\right)=\left(4x+4y+4-\left(x-y\right)\right)\left(4x+4y+4+x-y\right)\)
\(=\left(4x+4y+4\right)^2-\left(x-y\right)^2\le\left(4x+4y+4\right)^2\)
\(\Rightarrow VT\ge VP\)
"="<=>x=y=2