Question

[Lê Chí Cường]

Cho a,b,c > 0 thỏa mãn a² + b² + c² = 1 .

CMR:\(\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\le\frac{9}{2}\)

Answer 1

[Nguyễn Thiều Công Thành]

đặt \(A=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\)

\(\Rightarrow A-3=P=\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\)

áp dụng BĐT cô-si ta có:

\(a^2+b^2\ge2ab;b^2+c^2\ge2bc;c^2+a^2\ge2ca\)

\(\Rightarrow\frac{a^2+b^2}{2}\ge ab;\frac{b^2+c^2}{2}\ge bc;\frac{c^2+a^2}{2}\ge ca\)

\(\Rightarrow1-\frac{a^2+b^2}{2}\le1-ab;1-\frac{b^2+c^2}{2}\le1-bc;1-\frac{c^2+a^2}{2}\le1-ca\)

\(\Rightarrow P\le\frac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{2bc}{\left(a^2+b^2\right)+\left(a^2+c^2\right)}+\frac{2ca}{\left(a^2+b^2\right)+\left(b^2+c^2\right)}\)

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

Áp dụng BĐT Schwarts ta có:

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

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

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

\(\Rightarrow P\le\frac{1}{2}\left(\frac{a^2+b^2}{a^2+b^2}+\frac{b^2+c^2}{b^2+c^2}+\frac{c^2+a^2}{c^2+a^2}\right)=\frac{1}{2}.3=\frac{3}{2}\)

\(\Rightarrow P+3\le\frac{3}{2}+3\)

\(\Rightarrow A\le\frac{9}{2}\)

dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

Answer 2

[Kiệt Nguyễn]

Bất đẳng thức cần chứng minh tương đương: \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{-9}{2}\)

Theo bất đẳng thức Bunyakovsky dạng phân thức, ta được:  \(\frac{1}{ab-1}+\frac{1}{bc-1}+\frac{1}{ca-1}\ge\frac{9}{ab+bc+ca-3}\)

\(\ge\frac{9}{a^2+b^2+c^2-3}=\frac{9}{1-3}=\frac{-9}{2}\left(Q.E.D\right)\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)