[toán9] CM BĐT

H

hthtb22

Áp dụng bđt Côsi ta có

a2+12aa^2+1 \geq 2a

\Rightarrow a2+2b+32(a+b+1)a^2+2b+3 \geq2(a+b+1)

Ta có
A=aa2+2b+3+bb2+2c+3+cc2+2a+3\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{ c} {c^ 2+2a+3}

A\leq 12.(aa+b+1+bb+c+1+cc+a+1)\frac{1}{2}.(\frac{a}{a+b+1}+\frac{b}{b+c+1}+\frac{c} {c +a+1})

A \leq 12.(3b+1a+b+1+c+1b+c+1+a+1c+a+1)\frac{1}{2}.(3-\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{ a+1} {c +a+1})

A\leq 12.(3b+1a+b+1+c+1b+c+1+a+1c+a+1)\frac{1}{2}.(3-\frac{b+1}{a+b+1}+\frac{c+1}{b+c+1}+\frac{ a+1} {c +a+1})

A \leq 12.(3(b+1)2(a+b+1)(b+1)+(c+1)2(b+c+1)(c+1)+(a+1)2(c+a+1)(a+1))\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)})

A \leq 12.(3(a+b+c+3)2a2+b2+c2+ab+bc+ca+3a+3b+3c+3)\frac{1}{2}.(3-\frac{(a+b+c+3)^2}{a^2+b^2+c^2+ab+bc+ca+3a+3b+3c+3})

A\leq 12.(3(a+b+c+3)2ab+bc+ca+3a+3b+3c+6)\frac{1}{2}.(3-\frac{(a+b+c+3)^2}{ab+bc+ca+3a+3b+3c+6})

A \leq 12.(3(a+b+c+3)2(a+b+c)2a2b2c22+3a+3b+3c+6)\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})

A \leq 12.(32(a+b+c+3)2(a+b+c)23+6(a+b+c)+12)\frac{1}{2}.(3-\frac{2(a+b+c+3)^2}{(a+b+c)^2-3+6(a+b+c)+12})

A \leq 12.(32(a+b+c+3)2(a+b+c)2+6(a+b+c)+9)\frac{1}{2}.(3-\frac{2(a+b+c+3)^2}{(a+b+c)^2+6(a+b+c)+9})

A \leq 12[SIZE=4]\frac{1}{2}[SIZE=4]

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!
 
Top Bottom