Đáp án:
$GTNN$ $của$ $P$ $là$ $14$ $tại$ $x = 1; y = 0$
Giải thích các bước giải:
$P=\frac{2x + 8√x + 17}{√x + 2} + \frac{3y + 6√y + 5}{√y + 1}$
$P=\frac{(2x + 8√x + 8) + 9}{√x + 2} + \frac{(3y + 6√y + 3) + 2}{√y + 1}$
$P= \frac{2(x + 4√x + 4)+9}{√x+2} + \frac{3(y + 2√y + 1) + 2}{√y +1}$
$P = \frac{2(√x + 2)² + 9}{√x + 2} + \frac{3(√y + 1)² + 2}{√y +1}$
$Đặt$ $a = √x + 2 ; b = √y + 1$
$ĐK : a \geq 2 ; b \geq 1$ $thỏa$ $mãn$ $a+b \geq4$$(vì √x +√y \geq1)$
$Khi$ $đó$ $ta$ $có:$
$P=\frac{2a² + 9}{a} + \frac{3b² + 2}{b}$
$P=\frac{2a²}{a} + \frac{9}{a} + \frac{3b²}{b} +\frac{2}{b}$
$P = 2a + \frac{9}{a} + 3b + \frac{2}{b}$
$P = (a + b) + (a + \frac{9}{a}) + (2b + \frac{2}{b})$
$Theo$ $BĐT$ $Cauchy$ $ta$ $có:$
$a + \frac{9}{a} \geq 2√(a.\frac{9}{a}) = 2.3=6$
$2b + \frac{2}{b} \geq 2√(2b.\frac{2}{b}) = 2.2 =4$
$→ P \geq 4 + 6 + 4 = 14$
$Dấu$ $"="$ $xảy$ $ra$ $⇒$ $\left \{ {a + b = 4} \atop {a = \frac{9}{a} \atop {2b = \frac{2}{b}}}\right.⇒\left \{ {{a = 3} \atop {b = 1}} \right. ⇒\left \{ {{x = 1} \atop {y = 0}} \right.$
$Vậy$ $GTNN$ $của$ $P$ $là$ $14$ $tại$ $x = 1; y = 0$
