Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8

Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)

\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)

\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)

Lời giải:

Sửa: $x^2\geq y^2+z^2$
Áp dụng BĐT Cauchy-Schwarz:

$P\geq \frac{y^2+z^2}{x^2}+\frac{7x^2}{2}.\frac{4}{y^2+z^2}+2007$

$=\frac{y^2+z^2}{x^2}+\frac{14x^2}{y^2+z^2}+2007$

$=\frac{y^2+z^2}{x^2}+\frac{x^2}{y^2+z^2}+\frac{13x^2}{y^2+z^2}+2007$

$\geq 2+\frac{13x^2}{y^2+z^2}+2007$ (áp dụng BĐT Cô-si)

$\geq 2+13+2007=2022$ (do $x^2\geq y^2+z^2$)

Vậy $P_{\min}=2022$

\(E= {\sum {(yz)^2 \over xy+zx}}\)>=3/2 (AD BĐT Nesbit)

Dấu = xảy ra <=>x=y=z=1

đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow abc=\frac{1}{xyz}=1\)

Ta có : \(x+y=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=c\left(a+b\right)\)

Tương tự : \(y+z=a\left(b+c\right);x+z=b\left(c+a\right)\)

\(\Rightarrow E=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{3\sqrt[3]{abc}}{2}=\frac{3}{2}\)

\(\Rightarrow E\ge\frac{3}{2}\)

Vậy GTNN của E là \(\frac{3}{2}\Leftrightarrow x=y=z=1\)

1 cách khác Engel nữa,

\(E=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)

Áp dụng BĐT Bu-nhi-a-cốp-ski,ta có :

\(\left(a+b+c\right)^2=\left(\frac{a}{\sqrt{b+c}}.\sqrt{b+c}+\frac{b}{\sqrt{c+a}}.\sqrt{c+a}+\frac{c}{\sqrt{a+b}}.\sqrt{a+b}\right)^2\)

\(\le\left(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\right)\left(2a+2b+2c\right)\)

\(\Rightarrow E\ge\frac{a+b+c}{2}\ge\frac{3\sqrt[3]{abc}}{2}=\frac{3}{2}\)

Vậy ....

Ta có:

\(x^2+1=x^2+xy+yz+zx\)

           \(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)

Tương tự:

\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)

\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)

\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

TH1: x,y,z <0

\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)

TH2: x,y,z>0

\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)

Ta có \(1+z^2=xy+yz+zx+z^2\)

\(=y\left(x+z\right)+z\left(x+z\right)\)

\(=\left(x+z\right)\left(y+z\right)\)

CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)

Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)

\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)

 Tương tự như thế, ta được

\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)

 Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.