áp dụng BĐT $ \frac{(x+y)^2}{4} \geq xy$$=>abc \leq \frac{(a+b)^2}{4}.c=\frac{(6-c)^2}{4}.c=\frac{(6-c)}{4}[c(6-c)]\leq\frac{6-c}{4}.[\frac{(6-c+c)^2}{4}]=\frac{9}{4}(6-c)$Ta có $c \geq 3=> 6-c \leq 3=> đpcm$
áp dụng BĐT $ \frac{(x+y)^2}{4} \geq xy$$=>abc \leq \frac{(a+b)^2}{4}.c=\frac{(6-c)^2}{4}.c=\frac{(6-c)}{4}[c(6-c)]\leq\frac{6-c}{4}.[\frac{(6-c+c)^2}{4}]=\frac{9}{4}(6-c)$
áp dụng BĐT $ \frac{(x+y)^2}{4} \geq xy$$=>abc \leq \frac{(a+b)^2}{4}.c=\frac{(6-c)^2}{4}.c=\frac{(6-c)}{4}[c(6-c)]\leq\frac{6-c}{4}.[\frac{(6-c+c)^2}{4}]=\frac{9}{4}(6-c)$
Ta có $c \geq 3=> 6-c \leq 3=> đpcm$