Chủ đề này có 3 trả lời

[Arthur Pendragon]

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

$$\frac{5a^2-2}{b^2+bc+c^2}+\frac{5b^2-2}{c^2+ca+a^2}+\frac{5c^2-2}{a^2+ab+b^2} \geq 2$$

Bài viết đã được chỉnh sửa nội dung bởi tritanngo99: 04-02-2019 - 06:08

[DOTOANNANG]

Cho $\it{3}$ số dương $\it{a},\,\it{b},\,\it{c}$ thỏa mãn $\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}= \it{2}$ $\it{.}$ Chứng minh rằng $\it{:}$

$$\frac{\it{5}\,\it{a}^{\,\it{2}}- \it{2}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{5}\,\it{b}^{\,\it{2}}- \it{2}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{5}\,\it{c}^{\,\it{2}}- \it{2}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\geqq \it{2}$$

[DOTOANNANG]

Cho $\it{3}$ số dương $\it{a},\,\it{b},\,\it{c}$ thỏa mãn $\it{a}^{\,\it{2}}+ \it{b}^{\,\it{2}}+ \it{c}^{\,\it{2}}= \it{2}$ $\it{.}$ Chứng minh rằng $\it{:}$

$$\frac{\it{5}\,\it{a}^{\,\it{2}}- \it{2}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{5}\,\it{b}^{\,\it{2}}- \it{2}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{5}\,\it{c}^{\,\it{2}}- \it{2}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\geqq \it{2}$$

Bất đẳng thức trên sai $\it{!}$ Thật vậy $\it{:}$

Viết lại bất đẳng thức dưới dạng thuần nhất $\it{:}$

$$\frac{\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\geqq \it{2}$$

Ở $\it{2}$ vế $\it{,}$ nhân lên cùng $\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}$ được $\it{:}$

$\it{2}\,\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\leqq \it{bc}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}\,\,\it{)}+$

$$+ \it{ca}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{)}+ \it{ab}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}\,\,\it{)}$$

Ta có thể chứng minh bằng cách giả sử $\it{:}$ $\it{a}= \min \it{(}\,\,\it{a},\,\it{b},\,\it{c}\,\,\it{)}$ $\it{[}$ điều này không làm mất tính tổng quát trong chứng minh $\it{!}$ $\it{]}$ $\it{.}$ Khi đó $\it{:}$ $\it{b}- \it{a}= \it{p}\geqq \it{0}\,\,,\,\,\it{c}- \it{a}= \it{q}\geqq \it{0}\,\,\Rightarrow \,\,\it{b}= \it{a}+ \it{p}\,\,,\,\,\it{c}= \it{a}+ \it{q}$ $\it{,}$ $\it{[}$ dễ nhận thấy sẽ có một đa thức theo biến $\it{a}$ với hệ số của đa thức $\it{a}^{\,\it{0}}$ là không dương $\it{,}$ do đó ta phải phản chứng $\it{!}$ $\it{]}$ thay vào bất đẳng thức cần chứng minh $\it{:}$

$\it{2}\,\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\leqq$

$\leqq \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}\it{(}\,\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}\,\,\it{)}\it{[}\,\,\it{4}\,\it{a}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}\,\,\it{]}+$

$+ \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}\it{a}\it{(}\,\,\it{2}\,\it{a}+ \it{p}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}+ \it{q}\,\,\it{)}\it{[}\,\,\it{4}\it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{]}+$

$+ \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}\it{a}\it{(}\,\,\it{2}\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}+ \it{q}\,\,\it{)}\it{[}\,\,\it{4}\it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{]}$

$\Leftrightarrow \,\,\it{40}\,\it{a}^{\,\it{6}}+ \it{80}\it{(}\,\,\it{p}+ \it{q}\,\,\it{)}\it{a}^{\,\it{5}}+ \it{2}\it{(}\,\,\it{23}\,\it{p}^{\,\it{2}}+ \it{77}\,\it{pq}+ \it{23}\,\it{q}^{\,\it{2}}\,\,\it{)}\it{a}^{\,\it{4}}+ \it{8}\it{(}\,\,\it{p}^{\,\it{3}}+ \it{10}\,\it{pq}^{\,\it{2}}+ \it{10}\,\it{qp}^{\,\it{2}}+ \it{q}^{\,\it{3}}\,\,\it{)}\it{a}^{\,\it{3}}+ \it{4}\it{(}\,\,\it{3}\,\it{p}^{\,\it{3}}\it{q}+ \it{7}\,\it{p}^{\,\it{2}}\it{q}^{\,\it{2}}+ \it{3}\,\it{p}\it{q}^{\,\it{3}}\,\,\it{)}\it{a}^{\,\it{2}}\geqq \it{p}^{\,\it{2}}\it{q}^{\,\it{2}}\it{(}\,\,\it{p}^{\,\it{2}}+ \it{q}^{\,\it{2}}\,\,\it{)}$

Do đó bất đẳng thức trên sai với $\it{a}\rightarrow \it{0}^{\,\it{+}}\,\,,\,\,\it{p}\rightarrow \it{b}^{\,\it{-}}\,\,,\,\,\it{q}\rightarrow $$\it{c}^{\,\it{-}}$ $\it{.}$ Ta có bất đẳng thức của Vasile Cirtoaje như sau với $\it{3}$ số dương $\it{a},\,\it{b},\,\it{c}$ $\it{:}$

$$\frac{\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\leqq \it{3}$$

$\lceil$ B*W $\it{!}$   $\rfloor$ $\lceil$ Vasc $\it{'}$ s $\it{!}$$\rfloor$

[toanhoc2017]

Bất đẳng thức trên sai $\it{!}$ Thật vậy $\it{:}$

Viết lại bất đẳng thức dưới dạng thuần nhất $\it{:}$

$$\frac{\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\geqq \it{2}$$

Ở $\it{2}$ vế $\it{,}$ nhân lên cùng $\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}$ được $\it{:}$

$\it{2}\,\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\leqq \it{bc}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}\,\,\it{)}+$

$$+ \it{ca}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{)}+ \it{ab}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\it{(}\,\,\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}\,\,\it{)}$$

Ta có thể chứng minh bằng cách giả sử $\it{:}$ $\it{a}= \min \it{(}\,\,\it{a},\,\it{b},\,\it{c}\,\,\it{)}$ $\it{[}$ điều này không làm mất tính tổng quát trong chứng minh $\it{!}$ $\it{]}$ $\it{.}$ Khi đó $\it{:}$ $\it{b}- \it{a}= \it{p}\geqq \it{0}\,\,,\,\,\it{c}- \it{a}= \it{q}\geqq \it{0}\,\,\Rightarrow \,\,\it{b}= \it{a}+ \it{p}\,\,,\,\,\it{c}= \it{a}+ \it{q}$ $\it{,}$ $\it{[}$ dễ nhận thấy sẽ có một đa thức theo biến $\it{a}$ với hệ số của đa thức $\it{a}^{\,\it{0}}$ là không dương $\it{,}$ do đó ta phải phản chứng $\it{!}$ $\it{]}$ thay vào bất đẳng thức cần chứng minh $\it{:}$

$\it{2}\,\it{abc}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}\leqq$

$\leqq \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}\it{(}\,\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}\,\,\it{)}\it{[}\,\,\it{4}\,\it{a}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}\,\,\it{]}+$

$+ \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}\it{a}\it{(}\,\,\it{2}\,\it{a}+ \it{p}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}+ \it{q}\,\,\it{)}\it{[}\,\,\it{4}\it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{]}+$

$+ \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}\it{a}\it{(}\,\,\it{2}\,\it{a}+ \it{q}\,\,\it{)}\it{(}\,\,\it{2}\,\it{a}+ \it{p}+ \it{q}\,\,\it{)}\it{[}\,\,\it{4}\it{(}\,\,\it{a}+ \it{q}\,\,\it{)}^{\,\it{2}}- \it{(}\,\,\it{a}+ \it{p}\,\,\it{)}^{\,\it{2}}- \it{a}^{\,\it{2}}\,\,\it{]}$

$\Leftrightarrow \,\,\it{40}\,\it{a}^{\,\it{6}}+ \it{80}\it{(}\,\,\it{p}+ \it{q}\,\,\it{)}\it{a}^{\,\it{5}}+ \it{2}\it{(}\,\,\it{23}\,\it{p}^{\,\it{2}}+ \it{77}\,\it{pq}+ \it{23}\,\it{q}^{\,\it{2}}\,\,\it{)}\it{a}^{\,\it{4}}+ \it{8}\it{(}\,\,\it{p}^{\,\it{3}}+ \it{10}\,\it{pq}^{\,\it{2}}+ \it{10}\,\it{qp}^{\,\it{2}}+ \it{q}^{\,\it{3}}\,\,\it{)}\it{a}^{\,\it{3}}+ \it{4}\it{(}\,\,\it{3}\,\it{p}^{\,\it{3}}\it{q}+ \it{7}\,\it{p}^{\,\it{2}}\it{q}^{\,\it{2}}+ \it{3}\,\it{p}\it{q}^{\,\it{3}}\,\,\it{)}\it{a}^{\,\it{2}}\geqq \it{p}^{\,\it{2}}\it{q}^{\,\it{2}}\it{(}\,\,\it{p}^{\,\it{2}}+ \it{q}^{\,\it{2}}\,\,\it{)}$

Do đó bất đẳng thức trên sai với $\it{a}\rightarrow \it{0}^{\,\it{+}}\,\,,\,\,\it{p}\rightarrow \it{b}^{\,\it{-}}\,\,,\,\,\it{q}\rightarrow $$\it{c}^{\,\it{-}}$ $\it{.}$ Ta có bất đẳng thức của Vasile Cirtoaje như sau với $\it{3}$ số dương $\it{a},\,\it{b},\,\it{c}$ $\it{:}$

$$\frac{\it{4}\,\it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}}{\it{a}\it{(}\,\,\it{b}+ \it{c}\,\,\it{)}}+ \frac{\it{4}\,\it{b}^{\,\it{2}}- \it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}}{\it{b}\it{(}\,\,\it{c}+ \it{a}\,\,\it{)}}+ \frac{\it{4}\,\it{c}^{\,\it{2}}- \it{a}^{\,\it{2}}- \it{b}^{\,\it{2}}}{\it{c}\it{(}\,\,\it{a}+ \it{b}\,\,\it{)}}\leqq \it{3}$$

$\lceil$ B*W $\it{!}$   $\rfloor$ $\lceil$ Vasc $\it{'}$ s $\it{!}$$\rfloor$

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