[toán9] CM BĐT

H

hthtb22

Áp dụng bđt Côsi ta có

[tex]a^2+1 \geq 2a[/tex]

\Rightarrow [tex]a^2+2b+3 \geq2(a+b+1) [/tex]

Ta có
A=[tex]\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{ c} {c^ 2+2a+3}[/tex]

A\leq [tex]\frac{1}{2}.(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c} {c +a+1})[/tex]

A \leq [tex]\frac{1}{2}.(3-\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{ a+1} {c +a+1})[/tex]

A\leq [tex]\frac{1}{2}.(3-\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{ a+1} {c +a+1})[/tex]

A \leq [tex]\frac{1}{2}.(3-\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)})[/tex]

A \leq [tex]\frac{1}{2}.(3-\frac{(a+b+c+3)^2}{a^2+b^2+c^2+ab+bc+ca+3a+3b+3c+3})[/tex]

A\leq [tex]\frac{1}{2}.(3-\frac{(a+b+c+3)^2}{ab+bc+ca+3a+3b+3c+6})[/tex]

A \leq [tex]\frac{1}{2}.(3-\frac{(a+b+c+3)^2}{\frac{(a+b+c)^2-a^2-b^2-c^2}{2}+3a+3b+3c+6})[/tex]

A \leq [tex]\frac{1}{2}.(3-\frac{2(a+b+c+3)^2}{(a+b+c)^2-3+6(a+b+c)+12})[/tex]

A \leq [tex]\frac{1}{2}.(3-\frac{2(a+b+c+3)^2}{(a+b+c)^2+6(a+b+c)+9})[/tex]

A \leq [tex]\frac{1}{2}[SIZE=4][/tex]

Dấu = xảy ra \Leftrightarrow a=b=c=1
 
Last edited by a moderator:
V

vy000

có:

[TEX](a+b+c+3)^2[/TEX]

[TEX]=a^2+b^2+c^2+9+2ab+2bc+2ca+6a+6b+6c[/TEX]

[TEX]=12+2ab+2bc+2ca+6a+6b+6c[/TEX]

[TEX]=2(6+ab+bc+ca+3a+3b+3c)[/TEX]

[TEX]=2((a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1))[/TEX]

[TEX]\Rightarrow \frac{(a+b+c+3)^2}{(a+b+1)(b+1)+(b+c+1)(c+1)+(c+a+1)(a+1)}=2[/TEX]

[TEX]\Rightarrow 2 \leq \frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{a+1}{c+a+1}[/TEX]

[TEX]\Leftrightarrow 1 \geq \frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c}{c+a+1}[/TEX]

[TEX]\Leftrightarrow \frac{1}{2} \geq \frac{a}{2a+2b+21}+\frac{b}{2b+2c+2}+\frac{c}{2c+2a+2}[/TEX]

[TEX]\Leftrightarrow \frac{1}{2} \geq \frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}[/TEX]
 
B

bboy114crew

Bài này là bài toán của Phạm Kim Hùng!
Lời giải của nó khá phù hợp với THCS!
 
You must log in or register to reply here.
Top Bottom