cho a,b,c>0 có [tex]\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}[/tex] =1.CMR
a. $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\geq\frac{4}{3}$
b. $a+b+c+d\geq9(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d})$
Ta có: [tex]\frac{1}{1+a} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} \geq \frac{16}{a+b+c+d+4}[/tex]
Vậy a+b+c+d+a [tex]\leq 16[/tex]
Vậy a+b+c+d [tex]\leq 12[/tex]
a, Áp dụng BĐT Cosi ta có:
[tex]\frac{1}{a} + \frac{a}{9}[/tex] [tex]\geq \frac{2}{3}[/tex]
[tex]\frac{1}{b} + \frac{b}{9}[/tex] [tex]\geq \frac{2}{3}[/tex]
[tex]\frac{1}{c} + \frac{c}{9}[/tex] [tex]\geq \frac{2}{3}[/tex]
[tex]\frac{1}{d} + \frac{d}{9}[/tex] [tex]\geq \frac{2}{3}[/tex]
Vậy A + [tex]\frac{a+b+c+d}{9} \geq \frac{8}{3}[/tex]
Mà [tex]\frac{a+b+c+d}{9}[/tex] [tex]\leq \frac{4}{3}[/tex] nên A [tex]\geq \frac{4}{3}[/tex]