$A=4x^2+9y^2\\↔2A=2(4x^2+9y^2)\\↔2A=(1^2+1^2)(4x^2+9y^2)$
Áp dụng bất đẳng thức Bunhiacopxki
$→(1^2+1^2)(4x^2+9y^2)\ge (4x+9y)^2\\→2A\ge 169\\→A\ge \dfrac{169}{2}\\→\min A=\dfrac{169}{2}$
$→$ Dấu "=" xảy ra khi $\dfrac{1}{4x}=\dfrac{1}{9y}$
$↔4x=9y$ mà $4x+9y=13$
$→4x=9y=\dfrac{13}{2}$
$↔x=\dfrac{13}{8},y=\dfrac{13}{18}$
Vậy $\min A=\dfrac{169}{2}$ khi $x=\dfrac{13}{8},y=\dfrac{13}{18}$
