bat dang thuc
Cho
$\frac{a}{4}+\frac{b}{6}+\frac{c}{3} = 1
$Chứng minh rằng
:$\frac{(a+2b)^{3}}{5c+4a} + \frac{27c^{3}}{4a+4b+c} + \frac{(c+2a)^{3}}{a+2b+6c} \geqslant 16
$
Bất đẳng thức
bat dang thuc
Cho \frac{a}{4}+\frac{b}{6}+\frac{c}{3} = 1Chứng minh rằng \frac{(a+2b)^{3}}{5c+4a} + \frac{27c^{3}}{4a+4b+c} + \frac{(c+2a)^{3}}{a+2b+6c} \geqslant 16
GTLN, GTNN
bat dang thuc
Cho
$\frac{a}{4}+\frac{b}{6}+\frac{c}{3} = 1
$Chứng minh rằng
:$\frac{(a+2b)^{3}}{5c+4a} + \frac{27c^{3}}{4a+4b+c} + \frac{(c+2a)^{3}}{a+2b+6c} \geqslant 16
$
Bất đẳng thức