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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 \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=2\)
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
Bài 1:
\((x,y,z)=(\frac{2a^2}{bc}; \frac{2b^2}{ca}; \frac{2c^2}{ab})\) (\(a,b,c>0\) )
Khi đó:
\(\text{VT}=\frac{\frac{4a^4}{b^2c^2}}{\frac{4a^4}{b^2c^2}+\frac{4a^2}{bc}+1}+\frac{\frac{4b^4}{c^2a^2}}{\frac{4b^4}{c^2a^2}+\frac{4b^2}{ca}+4}+\frac{\frac{4c^4}{a^2b^2}}{\frac{4c^4}{a^2b^2}+\frac{4c^2}{ab}+4}\)
\(=\frac{a^4}{a^4+a^2bc+b^2c^2}+\frac{b^4}{b^4+b^2ac+a^2c^2}+\frac{c^4}{c^4+c^2ab+a^2b^2}\)
\(\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+(a^2b^2+b^2c^2+c^2a^2)}\)
(Áp dụng BĐT Cauchy_Schwarz)
Theo BĐT Cauchy dễ thấy:
\(a^2b^2+b^2c^2+c^2a^2\geq a^2bc+b^2ca+c^2ab\)
\(\Rightarrow \text{VT}\geq \frac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)}=\frac{(a^2+b^2+c^2)^2}{(a^2+b^2+c^2)^2}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=2$
Bài 2:
Đặt \((x,y,z)=\left(\frac{a}{b};\frac{b}{c}; \frac{c}{a}\right)\)
Ta có:
\(\text{VT}=\left(\frac{a}{b}+\frac{c}{b}-1\right)\left(\frac{b}{c}+\frac{a}{c}-1\right)\left(\frac{c}{a}+\frac{b}{a}-1\right)\)
\(=\frac{(a+c-b)(b+a-c)(c+b-a)}{abc}\)
Áp dụng BĐT Cauchy:
\((a+c-b)(b+a-c)\leq \left(\frac{a+c-b+b+a-c}{2}\right)^2=a^2\)
\((b+a-c)(c+b-a)\leq \left(\frac{b+a-c+c+b-a}{2}\right)^2=b^2\)
\((a+c-b)(c+b-a)\leq \left(\frac{a+c-b+c+b-a}{2}\right)^2=c^2\)
Nhân theo vế:
\(\Rightarrow [(a+c-b)(b+a-c)(c+b-a)]^2\leq (abc)^2\)
\(\Rightarrow (a+c-b)(b+a-c)(c+b-a)\leq abc\)
\(\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$ hay $x=y=z=1$
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