6 replies to this topic

[Daran Nguyen]

1. Cho a, b, c >0. Chứng minh rằng:

$\frac{a}{2a +b +c} + \frac{b}{a + 2b +c} + \frac{c}{a +b +2c} \leq  \frac{3}{4}$

 

2. Cho a, b, c > 0. Tìm Min:

$S = \frac{a^{3}}{b^{2}} + \frac{b^{3}}{c^{2}} + \frac{c^{3}}{a^{2}} + 27\left ( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} \right )$

 

3. Cho a, b > 0. Tìm Min:

$S = \frac{a +b }{\sqrt{ab}} + \frac{\sqrt{ab}}{ a +b }$

 

4. Cho $a \geq 2$ . Tìm Min:

$S = a + \frac{1}{a^{2}}$
 

5.  Cho a, b > 0; $a + b \leq  1$ .Tìm Min:

$S = \frac{1}{a^{2} + b ^{2}} + \frac{1}{ab} + 4ab$

[CaptainCuong]

 

4. Cho $a \geq 2$ . Tìm Min:

$S = a + \frac{1}{a^{2}}$
 

5.  Cho a, b > 0; $a + b \leq  1$ .Tìm Min:

$S = \frac{1}{a^{2} + b ^{2}} + \frac{1}{ab} + 4ab$

4.$\frac{1}{a^{2}}+\frac{a}{8}+\frac{a}{8}+\frac{3a}{4}\geq3\sqrt[3]{\frac{1}{64}}+\frac{3}{2}=\frac{3}{4}+\frac{3}{2}=\frac{9}{4}$

5.$\frac{1}{a^{2}+b^{2}}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\geq \frac{1}{(a+b)^{2}}+2\sqrt{4}-4(a+b)^{2}=1+4-4=1$

3.$a+b\geq 2\sqrt{ab}\Rightarrow \frac{a+b}{\sqrt{ab}}\geq 2\Rightarrow t\geq 2(\frac{a+b}{\sqrt{ab}}=t)$$t+\frac{1}{t}=\frac{t}{4}+\frac{1}{t}+\frac{3t}{4}\geq 2\sqrt{\frac{1}{4}}+\frac{3}{2}=1+\frac{3}{2}=\frac{5}{2}$

1.Áp dụng $\frac{4}{a+b}\leq \frac{1}{a}+\frac{1}{b}$

Edited by CaptainCuong, 03-12-2015 - 23:00.

[NTA1907]

1. Cho a, b, c >0. Chứng minh rằng:

$\frac{a}{2a +b +c} + \frac{b}{a + 2b +c} + \frac{c}{a +b +2c} \leq  \frac{3}{4}$

Ta có:

$\frac{a}{2a+b+c}\leq \frac{a}{4}(\frac{1}{a+b}+\frac{1}{a+c})=\frac{1}{4}(\frac{a}{a+b}+\frac{a}{a+c})$

Tương tự cộng lại ta được đpcm

[florairene]

2. Cho a, b, c > 0. Tìm Min:
$S = \frac{a^{3}}{b^{2}} + \frac{b^{3}}{c^{2}} + \frac{c^{3}}{a^{2}} + 27\left ( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} \right )$

$\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}}\geq 3\sqrt[3]{abc}$

$27(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca})\geq 27\cdot 3\frac{1}{\sqrt[3]{abc^{2}}}\Rightarrow S\geq 3\sqrt[3]{abc}+27\cdot 3\frac{1}{\sqrt[3]{abc^{2}}} =\frac{3}{2}\sqrt[3]{abc}+\frac{3}{2}\sqrt[3]{abc}+27\cdot 3\frac{1}{\sqrt[3]{abc^{2}}}=27\sqrt[3]{4}$

$Dau '=' xra\Leftrightarrow a=b=c=3\sqrt[3]{2}$

Không biết đúng ko nữa

Edited by florairene, 04-12-2015 - 14:19.

[Element hero Neos]

4.$\frac{1}{a^{2}}+\frac{a}{8}+\frac{a}{8}+\frac{3a}{4}\geq3\sqrt[3]{\frac{1}{64}}+\frac{3}{2}=\frac{3}{4}+\frac{3}{2}=\frac{9}{4}$

5.$\frac{1}{a^{2}+b^{2}}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\geq \frac{1}{(a+b)^{2}}+2\sqrt{4}-4(a+b)^{2}=1+4-4=1$

3.$a+b\geq 2\sqrt{ab}\Rightarrow \frac{a+b}{\sqrt{ab}}\geq 2\Rightarrow t\geq 2(\frac{a+b}{\sqrt{ab}}=t)$$t+\frac{1}{t}=\frac{t}{4}+\frac{1}{t}+\frac{3t}{4}\geq 2\sqrt{\frac{1}{4}}+\frac{3}{2}=1+\frac{3}{2}=\frac{5}{2}$

1.Áp dụng $\frac{4}{a+b}\leq \frac{1}{a}+\frac{1}{b}$

1.Dấu "=" xảy ra $\Leftrightarrow a=b$

3.Dấu "=" xảy ra $\Leftrightarrow a=b$

4.Dấu "=" xảy ra $\Leftrightarrow a=2$

5.Min =7 chứ không phải 1

[Element hero Neos]

Bài 5 ở đây có

[Daran Nguyen]

 

5. $\frac{1}{a^{2}+b^{2}}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\geq \frac{1}{(a+b)^{2}}+2\sqrt{4}-4(a+b)^{2}=1+4-4=1$

 

 

hình như số 1 đầu tiên phải là số 4 và số 4 tiếp theo là số 1 nhỉ ? 

Edited by Daran Nguyen, 04-12-2015 - 21:33.