3 replies to this topic

[marcoreus101]

1) Cho $a,b,c$ thực dương thỏa $a+b+c=1$. Chứng minh

$\sum \sqrt{\frac{a^2+2ab}{b^2+2c^2}}\geq \frac{1}{a^2+b^2+c^2}$

2) Cho $a,b,c$ thực dương thỏa $a+b+c=3$. Chứng minh

$\sum \frac{a}{b^3+ab}\geq 3$

[Hoang Tung 126]

1) Cho $a,b,c$ thực dương thỏa $a+b+c=1$. Chứng minh

$\sum \sqrt{\frac{a^2+2ab}{b^2+2c^2}}\geq \frac{1}{a^2+b^2+c^2}$

2) Cho $a,b,c$ thực dương thỏa $a+b+c=3$. Chứng minh

$\sum \frac{a}{b^3+ab}\geq 3$

Bài 1: Ta có : 

 

 $\sum \sqrt{\frac{a^2+2ab}{b^2+2c^2}}=\sum \frac{a^2+2ab}{\sqrt{(b^2+2c^2)(a^2+2ab)}}\geq \sum \frac{a^2+2ab}{\frac{a^2+2ab+b^2+2c^2}{2}}=2\sum \frac{a^2+2ab}{(a+b)^2+2c^2}\geq 2\sum \frac{a^2+2ab}{2(a^2+b^2)+2c^2}=\sum \frac{a^2+2ab}{a^2+b^2+c^2}=\frac{(\sum a)^2}{\sum a^2}=\frac{1}{\sum a^2}$

[Hoang Tung 126]

1) Cho $a,b,c$ thực dương thỏa $a+b+c=1$. Chứng minh

$\sum \sqrt{\frac{a^2+2ab}{b^2+2c^2}}\geq \frac{1}{a^2+b^2+c^2}$

2) Cho $a,b,c$ thực dương thỏa $a+b+c=3$. Chứng minh

$\sum \frac{a}{b^3+ab}\geq 3$

Bài 2: ta có :

 

 $\sum \frac{a}{b^3+ab}=\sum \frac{a}{b(a+b^2)}=\sum \frac{a+b^2-b^2}{b(a+b^2)}=\sum \frac{1}{b}-\sum \frac{b}{a+b^2}\geq \sum \frac{1}{b}-\sum \frac{b}{2\sqrt{ab^2}}=\sum \frac{1}{b}-\frac{1}{2}\sum \frac{1}{\sqrt{a}}\geq \sum \frac{1}{b}-\frac{1}{2}\sum (\frac{\frac{1}{a}+1}{2})=\sum \frac{1}{b}-\frac{1}{4}\sum \frac{1}{a}-\frac{3}{4}=\frac{3}{4}(\sum \frac{1}{a}-1)\geq \frac{3}{4}(\frac{9}{\sum a}-1)=\frac{3}{4}(\frac{9}{3}-1)=\frac{3}{2}$

[understand]

Bài 2: ta có :

 

 $\sum \frac{a}{b^3+ab}=\sum \frac{a}{b(a+b^2)}=\sum \frac{a+b^2-b^2}{b(a+b^2)}=\sum \frac{1}{b}-\sum \frac{b}{a+b^2}\geq \sum \frac{1}{b}-\sum \frac{b}{2\sqrt{ab^2}}=\sum \frac{1}{b}-\frac{1}{2}\sum \frac{1}{\sqrt{a}}\geq \sum \frac{1}{b}-\frac{1}{2}\sum (\frac{\frac{1}{a}+1}{2})=\sum \frac{1}{b}-\frac{1}{4}\sum \frac{1}{a}-\frac{3}{4}=\frac{3}{4}(\sum \frac{1}{a}-1)\geq \frac{3}{4}(\frac{9}{\sum a}-1)=\frac{3}{4}(\frac{9}{3}-1)=\frac{3}{2}$

bạn sai 1 lỗi rồi bạn ơi.3/4(9/1-1)=8