Áp Dụng BĐT svacxơ, ta có 

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}=9\left(ĐPCM\right)\)

^_^

Đặt a = \(x^2+2yz\); b = \(y^2+2xz\); c = \(z^2+2xy\)

\(\Rightarrow\)\(a,b,c>0\)và \(a+b+c=\left(x=y+z\right)^2=1\)

+) C/m : \(\left(a=b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Rightarrow\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\)

Hay \(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\)

\(\Rightarrow\)ĐPCM 

hên xui thôi -_-

CM BĐT phụ:  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Leftrightarrow3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\ge9\)(đúng) 

Áp dụng BĐT trên ta có: 

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=\frac{9}{\left(x+y+z\right)^2}=9\)

Cauchy - Schwarz dạng Engel :

\(\frac{1}{x^2+2xy}+\frac{1}{y^2+2yz}+\frac{1}{z^2+2zx}\ge\frac{\left(1+1+1\right)^2}{\left(x+y+z\right)^2}=9\)

Đẳng thức xảy ra <=> x = y = z = 1/3 

cảm ơn nha

Áp  dụng bđt Svac ta có:

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}=9\)

gg 

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+2yz+y^2+2xz+z^2+2xy}=\frac{9}{\left(x+y+z\right)^2}=9\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

Bạn tự c/m BĐT : \(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)

Dấu " = " xảy ra ta có:

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2zx}+\frac{1}{z^2+2xy}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2yz+2zx}+\frac{1}{z^2+2xy}\)\(\ge\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1}=9\)

Bạn tự giải dấu bằng nhé.

áp dụng bđt bunhia dạng phân thức ta có

\(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\)≥\(\frac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}\) =\(\frac{3^2}{\left(x+y+z\right)^2}\)=\(\frac{9}{1^2}\) =9

(đpcm) vậy dấu =xảy ra khi x=y=z=\(\frac{1}{3}\)

áp dụng bổ đề     \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)(bạn dùng cô-si,xét tích \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a+b+c\right)\))

\(\Leftrightarrow\frac{1}{x^2+2xy}+\frac{1}{y^2+2yz}+\frac{1}{z^2+2xz}\ge\frac{9}{\left(x+y+z\right)^2}=\frac{9}{1^2}\)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\right)[(x^2+2yz)+(y^2+2xz)+(z^2+2xy)]\geq (1+1+1)^2\)

\(\Leftrightarrow \frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\geq \frac{9}{x^2+2yz+y^2+2xz+z^2+2xy}=\frac{9}{(x+y+z)^2}=\frac{9}{3^2}=1\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z=1$