5 replies to this topic

[together1995]

1. Cho$ a,b,c>0$.CMR:$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \ge \frac{27}{2(a+b+c)^2}$.
2. Cho$ x,y,z>0$; $xy+xz+yz=1$. CMR: $\frac{x}{y.\sqrt{3}+z}+ \frac{y}{z.\sqrt{3}+x}+ \frac{z}{x.\sqrt{3}+y} \ge \frac{3.\sqrt{3}}{4}.$

[N H Tu prince]

1. Cho$ a,b,c>0$.CMR:$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \ge \frac{27}{2(a+b+c)^2}$.

$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{9}{a^2+b^2+c^2+ab+bc+ca}=\frac{9}{(a+b)^2+(b+c)^2+(c+a)^2}$
Áp dụng Cauchy-schwarz
$\frac{9}{(a+b)^2+(b+c)^2+(c+a)^2}\geq \frac{9}{\frac{1}{3}.2(a+b+c)^2}=\frac{27}{2(a+b+c)^2}$
Suy ra dpcm
Đẳng thức xảy ra khi a=b=c

Edited by hoangngocbao1997, 24-10-2012 - 18:17.

[Math Is Love]

1. Cho$ a,b,c>0$.CMR:$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \ge \frac{27}{2(a+b+c)^2}$.
2. Cho$ x,y,z>0$; $xy+xz+yz=1$. CMR: $\frac{x}{y.\sqrt{3}+z}+ \frac{y}{z.\sqrt{3}+x}+ \frac{z}{x.\sqrt{3}+y} \ge \frac{3.\sqrt{3}}{4}.$

Bài 2:(đề sai thì phải)
Áp dụng BĐT Cauchy-Schwarz,ta có:
$\sum \frac{x}{y.\sqrt{3}+z}=\sum \frac{x^2}{\sqrt{3}xy+zx}\geq \frac{(x+y+z)^2}{(\sqrt{3+1})(xy+yz+xz)}=\frac{(x+y+z)^2}{(\sqrt{3+1})}$
$\Rightarrow \sum \frac{x}{y.\sqrt{3}+z}\geq \frac{3}{\sqrt{3}+1}$
Mặt khác,ta có:
$(x+y+z)^2\geq 3(xy+yz+xz)=3$
Vậy $\Rightarrow \sum \frac{x}{y.\sqrt{3}+z}\geq \frac{3}{\sqrt{3}+1}$.
Dấu $"="$ khi $a=b=c=\frac{1}{\sqrt{3}}$

Edited by doxuantung97, 25-10-2012 - 10:05.

[phatthientai]

2. Cho$ x,y,z>0$; $xy+xz+yz=1$. CMR: $\frac{x}{y.\sqrt{3}+z}+ \frac{y}{z.\sqrt{3}+x}+ \frac{z}{x.\sqrt{3}+y} \ge \frac{3.\sqrt{3}}{4}.$

Bạn có thể xem lại đề không? Nếu $x=y=z=\frac{\sqrt{3}}{3}$ thì $$\frac{x}{y.\sqrt{3}+z}+\frac{y}{z.\sqrt{3}+x}+\frac{z}{x.\sqrt{3}+y}\ge \frac{3\sqrt{3}-3}{2}<\frac{3\sqrt{3}}{4}$$

[Secrets In Inequalities VP]

$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{9}{a^2+b^2+c^2+ab+bc+ca}=\frac{9}{(a+b)^2+(b+c)^2+(c+a)^2}$
Áp dụng Cauchy-schwarz
$\frac{9}{(a+b)^2+(b+c)^2+(c+a)^2}\geq \frac{9}{\frac{1}{3}.2(a+b+c)^2}=\frac{27}{2(a+b+c)^2}$
Suy ra dpcm
Đẳng thức xảy ra khi a=b=c

Sai rồi !

1. Cho$ a,b,c>0$.CMR:$ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)} \ge \frac{27}{2(a+b+c)^2}$.

Theo AM-GM : $abc\leq \frac{(a+b+c)^3}{27}$
$(a+b)(b+c)(c+a)\leq (\frac{2(a+b+c)}{3})^3= \frac{8(a+b+c)^3}{27}$
$\Rightarrow abc(a+b)(b+c)(c+a)\leq \frac{8(a+b+c)^6}{27.27}$
Lại theo AM-GM :
$VT\geq \frac{3}{\sqrt[3]{abc(a+b)(b+c)(c+a)}}\geq \frac{3}{\sqrt[3]{\frac{8(a+b+c)^6}{27.27}}}= \frac{27}{2(a+b+c)^2}$
$\Rightarrow Q.E.D$

[together1995]

Bài 2:(đề sai thì phải)
Áp dụng BĐT Cauchy-Schwarz,t$\Rightarrow \sum \frac{x}{y.\sqrt{3}+z}\geq \frac{3}{\sqrt{3}+1}$a có:
$\sum \frac{x}{y.\sqrt{3}+z}=\sum \frac{x^2}{\sqrt{3}xy+zx}\geq \frac{(x+y+z)^2}{(\sqrt{3+1})(xy+yz+xz)}=\frac{(x+y+z)^2}{(\sqrt{3+1})}$
Mặt khác,ta có:
$(x+y+z)^2\geq 3(xy+yz+xz)=3$
Vậy $\Rightarrow \sum \frac{x}{y.\sqrt{3}+z}\geq \frac{3}{\sqrt{3}+1}$.
Dấu $"="$ khi $a=b=c=\frac{1}{\sqrt{3}}$

Bài này trong sách đề là vậy, mình nghĩ nó cũng khá dễ, làm y như bạn đó, nhưng ko đc, chéc là do đề nhầm