Cho a, b, c, d > 0. Tìm GTNN của $P=\left ( 1+\frac{2a}{3b} \right )\left ( 1+\frac{2b}{3c} \right )\left ( 1+\frac{2c}{3d} \right )\left ( 1+\frac{2d}{3a} \right )$
Theo Holder thì :
$P\geq \left ( 1+\frac{2}{3} \right )^4=\frac{625}{81}$
Sử sụng Cauchy-Schwarz cũng được :
$P\geq \left ( 1+\frac{2}{3}\sqrt{\frac{a}{c}} \right )^2.\left ( 1+\frac{2}{3}\sqrt{\frac{c}{a}} \right )^2\geq \left ( 1+\frac{2}{3} \right )^4$
Hoặc dùng AM-GM bình thường ta có :
$P=\prod \left ( 1+\frac{2a}{3b} \right )=\prod \left ( \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{a}{3b}+\frac{a}{3b} \right )$
$\geq \prod \left ( \frac{5}{3}.\sqrt[5]{\frac{a^2}{b^2}} \right )=\frac{625}{81}$
Dấu "=" khi $a=b=c=d$