Question

cho a,b,c>0 sao cho a+b+c=3.tìm giá trị nhỏ nhất của biểu thức

P=\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{a^2+b^2+c^2}\)

Answer 1

\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a^2+b^2+c^2}\)

\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{9-2\left(ab+bc+ca\right)}\)

\(P=\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ca}+\frac{1}{9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(P\ge\frac{16}{3ab+3bc+3ca+9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{9}{ab+bc+ca}\right)\)

\(P\ge\frac{16}{9+ab+bc+ca}+\frac{6}{ab+bc+ca}\)

Sử dụng đánh giá quen thuộc:\(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)

\(\Rightarrow ab+bc+ca\le3\)

\(\Rightarrow P\ge\frac{16}{9+3}+\frac{6}{3}=2+\frac{4}{3}=\frac{10}{3}\)

"="<=>a=b=c=1