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 \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ba}\ge9\).
Áp dụng bất đẳng thức Cauchy - Schwarz ta có:
\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1^2}=9\)
Dấu "=" xảy ra khi: a = b = c = 1/3
Hoặc có thể dùng BĐT Cauchy như sau:
Ta có: \(\frac{1}{a^2+2bc}+9\left(a^2+2bc\right)\ge2\sqrt{\frac{1}{a^2+2bc}\cdot9\left(a^2+2bc\right)}=2\cdot3=6\)
Tương tự: \(\hept{\begin{cases}\frac{1}{b^2+2ca}+9\left(b^2+2ca\right)\ge6\\\frac{1}{c^2+2ab}+9\left(c^2+2ab\right)\ge6\end{cases}}\)
Cộng vế lại ta được:
\(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ca}+\frac{1}{c^2+2ab}+9\left(a^2+b^2+c^2+2ab+2bc+2ca\right)\ge18\)
\(\Rightarrow VT\ge18-9\left(a+b+c\right)^2\ge18-9=9\)
Dấu "=" xảy ra khi: a = b = c = 1/3
