\(A=\frac{xyz}{x+y}\Rightarrow\frac{1}{A}=\frac{x+y}{xyz}\)

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

 \(\frac{x+y}{xyz}=\frac{x}{xyz}+\frac{y}{xyz}=\frac{1}{yz}+\frac{1}{xz}\ge\frac{\left(1+1\right)^2}{yz+xz}=\frac{4}{z\left(x+y\right)}\)(1)

Lại có \(z\left(x+y\right)\le\frac{\left(x+y+z\right)^2}{4}=\frac{9}{4}\)(theo AM-GM) => \(\frac{4}{z\left(x+y\right)}\ge\frac{16}{9}\)(2)

Từ (1) và (2) => \(\frac{x+y}{xyz}\ge\frac{4}{z\left(x+y\right)}\ge\frac{16}{9}\)=> \(\frac{x+y}{xyz}\ge\frac{16}{9}\)hay \(\frac{1}{A}\ge\frac{16}{9}\)

=> A ≤ 9/16. Đẳng thức xảy ra <=> z = 3/2 ; x = y = 3/4

Vậy MaxA = 9/16 <=> x = y = 3/4 ; z = 3/2

\(9=3^2=\left(x+y+z\right)^2\ge4\left(x+y\right)z\)

\(\rightarrow9.\frac{x+y}{xyz}\ge4.\frac{\left(x+y\right)^2}{xy}\ge4.\frac{4xy}{xy}=16\)

\(\rightarrow\frac{x+y}{xyz}\ge\frac{16}{9}\rightarrow\frac{xyz}{x+y}\le\frac{9}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{3}{4};z=\frac{3}{2}\)

Áp dụng bđt : a^2+b^2+c^2 >= ab+bc+ca thì :

P = x^4+y^4+z^4/xyz >= x^2y^2+y^2z^2+z^2x^2/xyz

   >= xy.yz+yz.zx+zx.xy/xyz

     = xyz.(x+y+z)/xyz

     = x+y+z = -3

Dấu "=" xảy ra <=> x=y=z=-1 (T/m)

Vậy ...........

Tk mk nha

Lời giải:
Áp dụng BĐT Cauchy-Schwarz:

$A=\frac{1}{xz}+\frac{1}{xy}=\frac{1}{x}(\frac{1}{y}+\frac{1}{z})\geq \frac{1}{x}.\frac{4}{y+z}$

$=\frac{4}{x(y+z)}=\frac{4}{x(2-x)}$

Áp dụng BĐT AM-GM:

$x(2-x)\leq \left(\frac{x+2-x}{2}\right)^2=1$

$\Rightarrow A\geq \frac{4}{1}=4$
Vậy $A_{\min}=4$. Giá trị này đạt tại $x=1; y=z=\frac{1}{2}$

P=x/x+1 + y/y+1 + z/z+1=x+1-1/x+1 + y+1-1/y+1 + z+1-1/z+1

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

=3 - (1/x+1 + 1/y+1 + 1/z+1)

Áp dụng bđt cauchy- schwarz dạng engel:

1/x+1 + 1/y+1 + 1/z+1 = 1²/x+1 + 1²/y+1 + 1²/z+1 >/ (1+1+1)²/x+1+y+1+z+1 >/ 9/4 (do x+y+z=1)

=> P </ 3 - 9/4 = 3/4 

maxP=3/4 

We have:

\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)

Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)

Dau '=' xay ra khi \(x=y=z=1\)