Chủ đề này có 1 trả lời

[TanSan26]

Cho $a,b,c>0$ thỏa mãn: $a^2+b^2+c^2=1$. Chứng minh rằng:

$\sum \frac{1}{1-ab}\le \frac{9}{2}$

[Nguyenngoctu]

Cho $a,b,c>0$ thỏa mãn: $a^2+b^2+c^2=1$. Chứng minh rằng:

$\sum \frac{1}{1-ab}\le \frac{9}{2}$

Ta có: $$\begin{array}{l} \frac{1}{{1 - ab}} + \frac{1}{{1 - bc}} + \frac{1}{{1 - ca}} \le \frac{9}{2}\\ \Leftrightarrow \sum {\left( {\frac{1}{{1 - ab}} - 1} \right)} \le \frac{3}{2}\\ \Leftrightarrow \sum {\frac{{ab}}{{1 - ab}}} \le \frac{3}{2} \end{array}$$

$$\begin{array}{l} \sum {\frac{{ab}}{{1 - ab}}} = \sum {\frac{{2ab}}{{2 - 2ab}}} = \sum {\frac{{2ab}}{{{a^2} + {b^2} + 2{c^2} + {{\left( {a - b} \right)}^2}}}} \\ \le \sum {\frac{{2ab}}{{\left( {{a^2} + {c^2}} \right) + \left( {{b^2} + {c^2}} \right)}}} \le \frac{1}{2}\sum {\frac{{{{\left( {a + b} \right)}^2}}}{{\left( {{a^2} + {c^2}} \right) + \left( {{b^2} + {c^2}} \right)}}} \le \frac{1}{2}\sum {\left( {\frac{{{a^2}}}{{{a^2} + {c^2}}} + \frac{{{b^2}}}{{{b^2} + {c^2}}}} \right) = \frac{3}{2}} \end{array}$$