7 replies to this topic

[githenhi512]

1. Cho a,b,c>0. CM: $\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a}\geq 6$

2. Cho a,b,c>0 t/m: ab+bc+ca=3.CM: $\sum \frac{a}{2a^{2}+bc}\geq abc$

[NTA1907]

2. Cho a,b,c>0 t/m: ab+bc+ca=3.CM: $\sum \frac{a}{2a^{2}+bc}\geq abc$

Bđt$\Leftrightarrow \sum \frac{1}{bc(2a^{2}+bc)}\geq 1$

Áp dụng bđt Svac-sơ ta có:

$\sum \frac{1}{bc(2a^{2}+bc)}\geq \frac{9}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)}=\frac{9}{(ab+bc+ca)^{2}}=1$(đpcm)

Dấu = xảy ra$\Leftrightarrow a=b=c=1$

[githenhi512]

Bđt$\Leftrightarrow \sum \frac{1}{bc(2a^{2}+bc)}\geq 1$

Áp dụng bđt Svac-sơ ta có:

$\sum \frac{1}{bc(2a^{2}+bc)}\geq \frac{9}{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)}=\frac{9}{(ab+bc+ca)^{2}}=1$(đpcm)

Dấu = xảy ra$\Leftrightarrow a=b=c=1$

Bn giải thích chỗ này đc k? Mk k hiểu lắm.

[NTA1907]

Bn giải thích chỗ này đc k? Mk k hiểu lắm.

Nhân vào 2 vế $\frac{1}{abc}$

$\sum \frac{a}{2a^{2}+bc}=\sum \frac{a}{abc(2a^{2}+bc)}=\sum \frac{1}{bc(2a^{2}+bc)}$

[SKT T1 SPAK]

1 làm thế nào?

Edited by SKT T1 SPAK, 02-05-2016 - 16:58.

[PlanBbyFESN]

1. Cho a,b,c>0. CM: $\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a}\geq 6$

 

Cauchy-Schwarz:

 

$\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a}=\frac{(a+3c)^{2}}{(a+b)(a+3c)}+\frac{(c+3a)^{2}}{(b+c)(c+3a)}+\frac{(4b)^{2}}{4b(c+a)}\geq \frac{16(a+b+c)^{2}}{a^{2}+c^{2}+8bc+8ab+6ac}$

 

BĐT cần CM trở thành:

 

$\frac{16(a+b+c)^{2}}{a^{2}+c^{2}+8bc+8ab+6ac}\geq 6\Leftrightarrow 10a^{2}+10c^{2}+16b^{2}\geq 16bc+16ab+4ac$

 

$\Leftrightarrow 8(b-c)^{2}+8(a-b)^{2}+2(a-c)^{2}\geq 0$

......................................................................................

[cristianoronaldo]

1. Cho a,b,c>0. CM: $\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a}\geq 6$

2. Cho a,b,c>0 t/m: ab+bc+ca=3.CM: $\sum \frac{a}{2a^{2}+bc}\geq abc$

Bài 1:

Đặt a+b=x, b+c=y ,c+a=z thì ta được:a=$\frac{x+z-y}{2}$

                                                           b=$\frac{x+y-z}{2}$

                                                           c=$\frac{y+z-x}{2}$

Như vậy ta được:VT=$\frac{\frac{x+z-y}{2}+\frac{3(y+z-x)}{2}}{x}+\frac{\frac{y+z-x}{2}+\frac{3(x+z-y)}{2}}{y}+\frac{2(x+y-z)}{z}$

                                =$\frac{2z+y-x}{x}+\frac{2z+x-y}{y}+\frac{2x+2y-2z}{z}$

                                =$2.(\frac{x}{z}+\frac{z}{x})+2.(\frac{y}{z}+\frac{z}{y})+(\frac{x}{y}+\frac{y}{x})-4$

                                   $\geq 2.2+2.2+2-4=6$

Dấu ''='' xảy ra khi a=b=c

Edited by cristianoronaldo, 02-05-2016 - 20:56.

[dogamer01]

1. Cho a,b,c>0. CM: $\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a}\geq 6$

 

 

VT + 1 <=> $\frac{a+3c}{a+b}+\frac{c+3a}{b+c}+\frac{4b}{c+a} + 1\geq 7$

 

<=> $\frac{a+c}{a+b}+ \frac{2c}{a+b} +\frac{c+a}{b+c}+ \frac{2a}{b+c} + \frac{2b}{c+a} + \frac{b+a}{c+a} + \frac{b+c}{c+a} + \frac{2b}{c+a} \geq 7$

Nhận xét:

$\frac{a+c}{a+b}+ \frac{c+a}{b+c}+ \frac{b+a}{c+a} + \frac{b+c}{c+a} \geq 4$ (AM - GM 4 số)

$\frac{2c}{a+b} + \frac{2a}{b+c} + \frac{2b}{c+a} = 2 (\frac{c^{2}}{ac + cb} + \frac{a^{2}}{ab+ac} + \frac{b^{2}}{ba+bc}) \geq \frac{(a+b+c)^{2}}{ab + bc + ca} = 3$

 

=> VT $\geq 6$ (đpcm)