Chủ đề này có 3 trả lời

[hangnguyen2003]

Cho a,b,c>0. CMR:

$\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\leq \frac{a+b+c}{4}$

[DOTOANNANG]

$\it{2}\,\left ( \frac{\it{ab}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}+ \frac{\it{bc}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}}} \right )\geqq$

$\geqq \frac{\it{2}\,\left ( \it{a}^{\,\it{5}}+ \it{ba}^{\,\it{4}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}+ \it{bca}^{\,\it{3}}+ \it{5}\,\it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}+ \it{2}\,\it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}+ \it{b}$

Spoiler

$\sum\limits_{\it{cyc}} \frac{\it{2}\,\left ( \it{a}^{\,\it{5}}+ \it{ba}^{\,\it{4}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}+ \it{bca}^{\,\it{3}}+ \it{5}\,\it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}+ \it{2}\,\it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}= \it{0}$

[Song Binh]

Áp dụng bất đẳng thức phụ $\frac{1}{x+y}\leq \frac{1}{4}(\frac{1}{x}+\frac{1}{y})$

Ta có: $\frac{ab^2}{a^2+2b^2+c^2} = \frac{ab^2}{(a^2+b^2)+(b^2+c^2)} \leq \frac{ab^2}{4}(\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}) = \frac{1}{4}(\frac{ab^2}{a^2+b^2}+\frac{ab^2}{b^2+c^2}) \leq \frac{1}{4}(\frac{ab^2}{2ab}+\frac{ab^2}{2bc}) = \frac{1}{4}(\frac{b}{2}+\frac{ab}{2c})$

Tương tự ta có: $\frac{bc^2}{b^2+2c^2+a^2}\leq \frac{1}{4}(\frac{c}{2}+\frac{bc}{2a})$

                          $\frac{ca^2}{c^2+2a^2+b^2}\leq \frac{1}{4}(\frac{a}{2}+\frac{ac}{2b})$

Cộng theo vế của các bất đẳng thức ta có: 

        $\frac{ab^2}{a^2+2b^2+c^2} + \frac{bc^2}{b^2+2c^2+a^2} + \frac{ca^2}{c^2+2a^2+b^2} \leq \frac{1}{4}(\frac{b}{2}+\frac{ab}{2c}) + \frac{1}{4}(\frac{c}{2}+\frac{bc}{2a}) + \frac{1}{4}(\frac{a}{2}+\frac{ac}{2b}) = \frac{1}{4}[\frac{a+b+c}{2}+ \frac{1}{2}(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b})]$

Ta phải chứng minh: $\frac{ab}{c}+ \frac{bc}{a}+ \frac{ca}{b} \leq a+b+c$

(Nhưng mình chỉ chứng minh được $\frac{ab}{c}+ \frac{bc}{a}+ \frac{ca}{b} \geq a+b+c$ thôi........)

Bài viết đã được chỉnh sửa nội dung bởi Song Binh: 23-12-2018 - 15:32

[DOTOANNANG]

$\it{2}\,\left ( \frac{\it{ab}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}+ \frac{\it{bc}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}}} \right )\geqq$

$\geqq \frac{\it{2}\,\left ( \it{a}^{\,\it{5}}+ \it{ba}^{\,\it{4}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}+ \it{bca}^{\,\it{3}}+ \it{5}\,\it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}+ \it{2}\,\it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}+ \it{b}$

Spoiler

$\sum\limits_{\it{cyc}} \frac{\it{2}\,\left ( \it{a}^{\,\it{5}}+ \it{ba}^{\,\it{4}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}+ \it{bca}^{\,\it{3}}+ \it{5}\,\it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}+ \it{2}\,\it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}= \it{0}$

Spoiler

$\sum\limits_{\it{cyc}} \frac{\it{2}\,\left ( \it{a}^{\,\it{5}}+ \it{ba}^{\,\it{4}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}+ \it{bca}^{\,\it{3}}+ \it{5}\,\it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}+ \it{2}\,\it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}= \it{0}$

Với $\it{a}= \it{b}- \it{1}= \it{c}- \it{2}= \it{1}\Rightarrow \it{0}\neq \it{0}$ (vô lí!)

 

Đúng là,

 

$\it{2}\,\left ( \frac{\it{ab}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}+ \frac{\it{bc}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}}} \right )\leqq$

$\leqq \frac{-\,\it{a}^{\,\it{5}}+ \it{5}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}- \it{bca}^{\,\it{3}}- \it{a}^{\,\it{2}}\it{b}^{\,\it{3}}+ \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}+ \it{4}\,\it{ab}^{\,\it{4}}+ \it{cab}^{\,\it{3}}+ \it{b}^{\,\it{5}}- \it{cb}^{\,\it{4}}}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}} \right )}+$

$+ \frac{\it{2}\left ( \it{a}^{\,\it{5}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{3}}+ \it{bca}^{\,\it{3}}+ \it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}- \it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}+ b$

 

và,

 

$\sum\limits_{\it{cyc}} \frac{-\,\it{a}^{\,\it{5}}+ \it{5}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{2}}- \it{bca}^{\,\it{3}}- \it{a}^{\,\it{2}}\it{b}^{\,\it{3}}+ \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}+ \it{4}\,\it{ab}^{\,\it{4}}+ \it{cab}^{\,\it{3}}+ \it{b}^{\,\it{5}}- \it{cb}^{\,\it{4}}}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}} \right )}=$

$=-\,\sum\limits_{\it{cyc}} \frac{\it{2}\left ( \it{a}^{\,\it{5}}- \it{4}\,\it{a}^{\,\it{3}}\it{b}^{\,\it{3}}+ \it{bca}^{\,\it{3}}+ \it{a}^{\,\it{2}}\it{b}^{\,\it{3}}- \it{ca}^{\,\it{2}}\it{b}^{\,\it{2}}- \it{ab}^{\,\it{4}}- \it{cab}^{\,\it{3}}- \it{b}^{\,\it{5}}+ \it{cb}^{\,\it{4}} \right )}{\left ( \it{a}^{\,\it{2}}+ \it{3}\,\it{b}^{\,\it{2}} \right )\left ( \it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{2}\,\it{c}^{\,\it{2}} \right )}$

 

$$\sum\limits_{\it{cyc}} \frac{\it{ab}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{2}\,\it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}\leqq \frac{\it{1}}{\it{16}}\left ( \sum\limits_{\it{cyc}} \frac{\it{9}\,\it{ab}^{\,\it{2}}}{\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}}+ \sum\limits_{\it{cyc}} \frac{\it{ab}^{\,\it{2}}}{\it{b}^{\,\it{2}}} \right )\leqq \frac{\it{a}+ \it{b}+ \it{c}}{\it{4}}$$

 

$\lceil$ Bổ đề Titu $\rfloor$ $\lceil\,\,\it{3}\,\sum\limits_{\it{cyc}} \it{ab}^{\,\it{2}}\leqq \sum\limits_{\it{cyc}} \it{a}\sum\limits_{\it{cyc}} \it{a}^{\,\it{2}}\Leftrightarrow \sum\limits_{\it{cyc}} \it{a}\left ( \it{a}- \it{c} \right )^{\,\it{2}}\geqq \it{0}\,\,\rfloor$