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\(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=1\\xyz\left(x+y+z\right)\left(x+1\right)\left(y+1\right)\left(z+1\right)=1296\end{matrix}\right.\)
Đặt \(\dfrac{1}{x+1}=a;\dfrac{1}{y+1}=b;\dfrac{1}{z+1}=c\left(a,b,c>0\right)\)
\(\Rightarrow a+b+c=1\)
\(\dfrac{1}{x+1}=a\)
\(\Rightarrow x+1=\dfrac{1}{a}\)
\(\Rightarrow x=\dfrac{1}{a}-1=\dfrac{1-a}{a}=\dfrac{b+c}{a}\)
Tương tự, ta có: \(y=\dfrac{a+c}{b};z=\dfrac{a+b}{c}\)
Đặt \(M=xyz\left(x+y+z\right)\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(=\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{abc}\times\left(\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\right)\times\dfrac{1}{abc}\)
\(=\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{a^2b^2c^2}\times\left(\dfrac{b}{a}+\dfrac{a}{b}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{c}\right)\)
\(\ge\dfrac{8abc}{a^2b^2c^2}\times\left(2+2+2\right)\) (bđt AM - GM)
\(\ge\dfrac{8}{\dfrac{\left(a+b+c\right)^3}{27}}\times6=1296\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=2\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{1}{z}=-\left(\frac{1}{x}+\frac{1}{y}\right)\\ \frac{2}{xy}-\frac{1}{z^2}=4\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} \frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\\ \frac{2}{xy}-\frac{1}{z^2}=4\end{matrix}\right.\)
\(\Rightarrow \frac{2}{xy}-\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\right)=4\)
\(\Leftrightarrow -\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=4>0\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}< 0\) (vô lý)
Do đó không tồn tại $x,y,z$ kéo theo không tồn tại giá trị của P
1 + y2 = xy + yz + xz + y2 = (x + y)(y + z)
1 + z2 = xy + yz + xz + z2 = (x + z)(z + y)
1 + x2 = xy + yz + xz + x2 = (y + x)(x + z)
Sau khi thay vào và rút gọn ta được
S = x(y + z) + y(x + z) + z(x + y)
S = 2(xy + yz + xz) = 2.1 = 2
Ta có \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xyz}=1\)
\(\Leftrightarrow\dfrac{\left(yz\right)^2+\left(xz\right)^2+\left(xy\right)^2+2xyz}{\left(xyz\right)^2}=1\)
<=> (xy)2 + (yz)2 + (zx)2 + 2xyz = (xyz)2
<=> (xy)2 + (yz)2 + (xz)2 + 2xyz(x + y + z) = (xyz)2
<=> (xy + yz + zx)2 = (xyz)2
<=> \(\left[{}\begin{matrix}xy+yz+zx=xyz\\xy+yz+zx=-xyz\end{matrix}\right.\)
+) Khi xy + yz + zx = -xyz
=> \(\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=-1< 0\left(\text{loại}\right)\)
=> xy + yz + zx = xyz
<=> \(xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=xyz\Leftrightarrow xyz\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1\right)=0\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\)
<=> \(\dfrac{x+y}{xy}=\dfrac{-\left(x+y\right)}{\left(x+y+z\right)z}\)
<=> \(\left(x+y\right)\left(\dfrac{1}{xz+yz+z^2}+\dfrac{1}{xy}\right)=0\)
<=> \(\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{\left(zx+yz+z^2\right)xy}=0\)
<=> \(\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
Khi x = -y => y = 1 => P = 1
Tương tự y = -z ; z = -x được P = 1
Vậy P = 1
tks b nha