Question

[pham trung thanh]

Cho a;b;c > 0 và \(a^2+b^2+c^2=1\). CMR:

              \(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ca}< \frac{3}{2}\)

Answer 1

[Le Van Hung]

Đặt A=\(\frac{1}{3-ab}+\frac{1}{3-bc}+\frac{1}{3-ac}\)

      \(\Rightarrow4A=\frac{4}{3-ab}+\frac{4}{3-bc}+\frac{4}{3-ac}\)

 Ap dung BĐT cauchy-schawst ta co

 4A\(\le\frac{1}{2}+\frac{1}{1-ab}+\frac{1}{2}+\frac{1}{1-bc}+\frac{1}{2}+\frac{1}{1-ac}\)

4A-3\(\le\frac{3}{2}+\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ac}{1-ac}\)

Lai co \(ab\le\frac{a^2+c^2}{2}\)

\(1-ab\le\frac{2-a^2-c^2}{2}\)

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

CMTT .................................( bạn tự chứng minh nhé)

\(\Rightarrow\text{4a-3}\le\frac{3}{2}+\frac{1}{2}\left[\frac{\left(a+b\right)^2}{\left(a^2+b^2\right)\left(b^2+c^2\right)}+\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)\left(a^2+c^2\right)}+\frac{\left(a+c\right)^2}{\left(b^2+c^2\right)\left(a^2+c^2\right)}\right]\)

tiep tuc ap dung BĐT cauchy-schwast ta co

\(4A-3\le\frac{3}{2}-\frac{1}{2}+3\)

\(\Leftrightarrow A\le\frac{3}{2}\)

dau "=" xay ra khi

1-ab=2

1-bc=2( vô lí)

1-ac=2

Vay khong xay ra dau "="

k cho minh nhe

Answer 2

[alibaba nguyễn]

Đề sai khỏi làm