CM BĐT băng pp đặt ẩn phụ

D

darkangel98

Đặt:
$a=\dfrac{ x+y}{2x} \\
b=\dfrac{ y+z}{2y} \\
c=\dfrac{ x+z}{2z} \\
a+b+c=\dfrac{ 3}{2}+\dfrac{ y}{2x}+\dfrac{ x}{2z}+\dfrac{ z}{2y} \ge 3 $

Ta có:

$VT=\dfrac{ 1}{\sqrt{ a}}+\dfrac{ 1}{\sqrt{ b}}+\dfrac{ 1}{\sqrt{ c}} \le \sqrt{ 3.(\dfrac{ 1}{a}+\dfrac{ 1}{b}+\dfrac{ 1}{c})} \le 3$

mình chưa hiểu bước cuối cùng lắm.Bạn gthích được không?
 
V

vuive_yeudoi

Dùng Cauchy Schwarz
$$ \left( \sum \sqrt{\frac{x}{x+y}} \right)^2 \le \left( \sum \left(x+z \right) \right) \cdot \left( \sum \frac{x}{ \left( x+y \right) \left(x+z \right)} \right) $$
Ta thấy
$$ \left( \sum \left(x+z \right) \right) \cdot \left( \sum \frac{x}{ \left( x+y \right) \left(x+z \right)} \right)=\frac{4 \left( x+y+z \right) \left( xy+yz+zx \right)}{ \left( x+y \right) \left(y+z \right) \left( z+x \right)} $$
Cần chứng minh
$$ \frac{4 \left( x+y+z \right) \left( xy+yz+zx \right)}{ \left( x+y \right) \left(y+z \right) \left( z+x \right)} \le \frac{9}{2}$$
Chuyện đó đúng bởi vì
$$ \frac{9}{2}-\left(\frac{4 \left( x+y+z \right) \left( xy+yz+zx \right)}{ \left( x+y \right) \left(y+z \right) \left( z+x \right)} \right)=\frac{z \left( x-y \right)^2+x \left( y-z \right)^2 + y \left(z-x \right)^2}{2 \left( x+y \right) \left(y+z \right) \left( z+x \right)} \ge 0$$
 
D

darkangel98

Cm bdt

Cho a,b,c>0; a^2+b^2+c^2=1 .CMR
[TEX]\frac{a+b}{1-ab}+\frac{b+c}{1-bc}+\frac{c+a}{1-ca}\leq3(a+b+c)[/TEX]
 
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