Chủ đề này có 3 trả lời

[minh hai nguyen]

Cho a,b,c là các số dương thỏa mãn $a^{2}+b^{2}+c^{2}=3$

Chứng minh rằng: $\frac{a}{a^{2}+2b+3}+\frac{b}{b^{2}+2c+3}+\frac{c}{c^{2}+2a+3}\leq \frac{1}{2}$

 

 

Bài viết đã được chỉnh sửa nội dung bởi phamngochung9a: 31-05-2016 - 12:19

[ngothithuynhan100620]

Ta có: $a^{2}+1\ge 2a;b^{2}+1\ge 2b;c^{2}+1\ge 2c$

$=>\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\le\frac{a}{2\left(a+b+1\right)}+\frac{b}{2\left(b+c+1\right)}+\frac{c}{2\left(c+a+1\right)}$

Ta đi CM: $\frac{a}{2\left(a+b+1\right)}+\frac{b}{2\left(b+c+1\right)}+\frac{c}{2\left(c+a+1\right)}\le\frac{1}{2}$

$\iff \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}\le 1$

$\iff 1-\frac{a}{a+b+1}+1-\frac{b}{b+c+1}+1-\frac{c}{c+a+1}\ge 2$

$\iff \frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}\ge 2$

Ta có: $\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}=\frac{(b+1)^2}{(a+b+1)(b+1)}+\frac{(c+1)^2}{(b+c+1)(c+1)}+\frac{(a+1)^2}{(c+a+1)(a+1)}\ge\frac{(a+b+c+3)^2}{(a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1)}$

Mà: $\frac{(a+b+c+3)^2}{(a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1)}=\frac{(a+b+c+3)^2}{ab+bc+ca+3(a+b+c)+a^{2}+b^{2}+c^{2}+3}=\frac{(a+b+c+3)^2}{\frac{(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})}{2}+3(a+b+c)+a^{2}+b^{2}+c^{2}+3}=\frac{(t+3)^2}{\frac{t^{2}-3}{2}+3t+3+3}=\frac{(t+3)^2}{\frac{t^{2}}{2}+3t+\frac{9}{2}}=2  (t=a+b+c) $

=> Dpcm. Dau = xay ra khi a=b=c=1. 

Bài viết đã được chỉnh sửa nội dung bởi ngothithuynhan100620: 20-04-2016 - 08:08

[ngocminhxd]

bất đẳng thức được sử dụng này là gì ạ

Ta có: $\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}=\frac{(b+1)^2}{(a+b+1)(b+1)}+\frac{(c+1)^2}{(b+c+1)(c+1)}+\frac{(a+1)^2}{(c+a+1)(a+1)}\ge\frac{(a+b+c+3)^2}{(a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1)}$
 

[MinMax2k]

Ta có: $a^2+2b+1=a^2+1+2b+2\geq 2a+2b+2$

$\Rightarrow P\leq \frac{1}{2}(\Sigma (\frac{a}{a+b+1})$

$\Rightarrow \frac{3}{2}-P\geq$ $\frac{1}{2}$$(\Sigma \frac{b+1}{a+b+1})$

A/d Cauchy-Schwarz ta được:

$Q=\Sigma \frac{b+1}{a+b+1}=\Sigma \frac{(b+1)^2}{(b+1)(a+b+1)}\geq \frac{(a+b+c+3)^2}{\Sigma (a+1)(a+c+1)}$

Ta có:

$\Sigma (a+1)(a+c+1)=(a^2+b^2+c^2+ab+bc+ca+3(a+b+c)+3=\frac{1}{2}(a^2+b^2+c^2)+ab+bc+ca+3(a+b+c)+\frac{9}{2}=\frac{1}{2}(a+b+c+3)^2$

Suy ra:

$Q\geq \frac{(a+b+c+3)^2}{\frac{1}{2}(a+b+c+3)^2}\Rightarrow .............$

$\Rightarrow P\leq \frac{1}{2}$