Bđt <=> $(\left ( \frac{2}{\sqrt{\frac{1+a^{2}}{2}}+\frac{2a}{1+a}} \right )^{\frac{1}{3}}+\left ( \frac{2}{\sqrt{\frac{1+b^{2}}{2}}+\frac{2b}{1+b}} \right )^{\frac{1}{3}}+\left ( \frac{2}{\sqrt{\frac{1+c^{2}}{2}}+\frac{2c}{1+c}} \right )^{\frac{1}{3}})^{3}\leq 27$
Áp dụng bất đẳng thức Holder, ta có:
$(\left ( \frac{2}{\sqrt{\frac{1+a^{2}}{2}}+\frac{2a}{1+a}} \right )^{\frac{1}{3}}+\left ( \frac{2}{\sqrt{\frac{1+b^{2}}{2}}+\frac{2b}{1+b}} \right )^{\frac{1}{3}}+\left ( \frac{2}{\sqrt{\frac{1+c^{2}}{2}}+\frac{2c}{1+c}} \right )^{\frac{1}{3}})^{3}\leq (1+1+1)(1+1+1)\sum \frac{2}{\sqrt{\frac{1+a^{2}}{2}}+\frac{2a}{1+a}}\leq 9\sum \frac{2}{\frac{1+a}{2}+\frac{2a}{1+a}}$
Ta cần chứng minh: $\sum \frac{2}{\frac{1+a}{2}+\frac{2a}{1+a}}\leq 3$
Thật vậy sử dụng bổ đề quen thuộc sao: a>0, b>0 thì $\frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}$
=> $\sum \frac{2}{\frac{1+a}{2}+\frac{2a}{1+a}}\leq \frac{2}{4}\sum (\frac{2}{1+a}+\frac{1+a}{2a})=\frac{1}{2}\sum (\frac{a^{2}+6a+1}{2a+2a^{2}})$
Ta chỉ cần chứng minh $\frac{1}{2}\sum (\frac{a^{2}+6a+1}{2a+2a^{2}})\leq 3$
$<=> 2\sum \frac{a^{2}+6a+1}{2a+2a^{2}}\leq 3.4=12$
$<=> \frac{2a^{2}+12a+2}{2a+2a^{2}}-1+\frac{2b^{2}+12b+2}{2b+2b^{2}}-1+\frac{2c^{2}+12c+2}{2c+2c^{2}}-1\leq 9$
$<=> \sum \frac{5a+1}{a^{2}+a}\leq 9$
<=> $\frac{5a+1}{a^{2}+a}-5+ \frac{5b+1}{b^{2}+b}-5+ \frac{5c+1}{c^{2}+c}-5\leq -6$
$<=> \sum \frac{1-5a^{2}}{a^{2}+a}\leq -6$
Bài viết đã được chỉnh sửa nội dung bởi Duy Thai2002: 15-10-2017 - 11:57