Question

Cho a,b,c là 3 số dương thỏa mãn a+b+c=1. CMR

a,1/b+c + 1/c+a + 1/a+b >4

b, b+2b+c>= 4(1-a)(1-b)(1-c)

Answer 1

giúp mk vs plzzz

Answer 2

\(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{9}{2\left(a+b+c\right)}=\frac{9}{2}>4\)

Câu b chắc là \(a+2b+c\ge4\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

BĐT tương đương:

\(a+2b+c\ge4\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Ta có:

\(VP=4\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\left(a+2b+c\right)^2\left(c+a\right)\)

\(VP\le\left(a+2b+c\right)\left(a+2b+c\right)\left(c+a\right)\le\frac{1}{4}\left(a+2b+c\right)\left(a+2b+c+c+a\right)^2\)

\(\Rightarrow VP\le\frac{1}{4}\left(a+2b+c\right)\left(2a+2b+2c\right)^2=a+2b+c\) (đpcm)

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