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 $x=y=z=1$

a) Áp dụng bất đẳng thức Cauchy-Schwarz , ta được
\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}=1\)(đpcm)

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: 

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

Áp dụng BĐT Cauchy-schwarz dạng engel,ta có:

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

\(\Rightarrowđpcm\)

Áp dụng bất đẳng thức Cauchy-Schwarz,ta có:

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

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

( áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\))

Dấu "=" xảy ra khi x=y=z=1

Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)

Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)

Tương tự thay vào mà quy đồng

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0< =>\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=0< =>xy+yz+zx=0\)

Khi đó : \(x^2+2yz=x^2+2yz-xy-yz-zx=x^2-xy+yz-zx=\left(x-z\right)\left(x-y\right)\)

Bằng phép chứng minh tương tự ta được : \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)

Đặt \(A=\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

\(< =>-A=\frac{x^2}{\left(x-y\right)\left(z-x\right)}+\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(y-z\right)}\)

\(=\frac{x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=...\)đến đây nhân tung rồi ghép cặp sẽ ra kq = 1 thì phải 

làm luôn đỡ lòng vòng :(

\(=\frac{x^2\left(y-z\right)+y^2z-y^2x+z^2x-z^2y}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{x^2\left(y-z\right)+zy\left(y-z\right)-x\left(y^2-z^2\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

\(=\frac{x^2\left(y-z\right)+zy\left(y-z\right)-x\left(y-z\right)\left(y+z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(y-z\right)\left(x^2+zy-xy-xz\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

\(=\frac{\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=-1\)

\(< =>-A=-1< =>A=1\)

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=\(\frac{1}{3}\)