Question

[Không Cần Biết 2]

Với \(a+b+c\le1\) và a, b, c >0

CMR:\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ba}\ge9\)

Answer 1

[nghiem dinh hieu]

sai de thi phai

Answer 2

[The Angry]

ĐK:\(a+b+c\le1|a,b,c>0\)

Chỉ có TH \(a=b=c=\frac{1}{3}\)\(\Rightarrow TH:a+b+c=1\)

\(\Rightarrow\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}+\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}+\frac{1}{\left(\frac{1}{3}\right)^2+2.\frac{1}{3}.\frac{1}{3}}\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2+2\left(\frac{1}{3}\right)^2}3\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2\left(2+1\right)}3\ge9\)\(=\frac{1}{\left(\frac{1}{3}\right)^2.3}3\ge9\)\(=\frac{1}{\frac{1}{3}.\frac{1}{3}.3}3\ge9\)\(=\frac{1}{\frac{1}{3}}3\ge9\)\(=\frac{3}{\frac{1}{3}}\ge9\)\(=3:1:3\ge9\)\(=1\ge9\)( loại )

Vậy không thể CMR \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ba}\ge9\).