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\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)
=>\(8xyz=xyz+\sum x+\sum xy+1\)
=>\(\sum x^2+14xyz=\left(\sum x\right)^2+2\sum x+2\)
mặt khác
\(8=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge\dfrac{8}{\sqrt[3]{xyz}}\rightarrow xyz\ge1\)
đặt \(\sum x=a\left(a\ge3\right)\)
khi đó \(P=\dfrac{a^2+2a+2}{4a^2+15xyz}\le\dfrac{a^2+2a+2}{4a^2+15}\)
\(\dfrac{a^2+2a+2}{4a^2+15}=\dfrac{1}{3}-\dfrac{\left(a-3\right)^2}{12a^2+45}\le\dfrac{1}{3}\)
vậy max bằng 1/3 khi x=y=z=1
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+\dfrac{2}{xy}+\dfrac{2}{yz}+\dfrac{2}{xz}-\dfrac{2}{xy}-\dfrac{2}{yz}-\dfrac{2}{xz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2.\dfrac{x+y+z}{xyz}}=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\text{|}\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\text{|}\)
\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}+2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)-2\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{z}{xyz}+\dfrac{y}{xyz}+\dfrac{x}{xyz}\right)}\\ =\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
Áp dụng bđt Cauchy-Schwarz:
\(A=\dfrac{1}{\sqrt{x\left(y+2z\right)}}+\dfrac{1}{\sqrt{y\left(z+2x\right)}}+\dfrac{1}{\sqrt{z\left(x+2y\right)}}\)
\(\ge\dfrac{\left(1+1+1\right)^2}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
\(=\dfrac{9}{\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}}\)
Áp dụng liên tiếp Bunyakovsky và AM-GM:
\(\left(\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+\sqrt{z\left(x+2y\right)}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left[x\left(y+2z\right)+y\left(z+2x\right)+z\left(x+2y\right)\right]\)
\(=3.3\left(xy+yz+xz\right)\)
Mà \(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=3\)
\(3.3\left(xy+yz+xz\right)\le3.3=9\)
\(\Leftrightarrow\sqrt{x\left(y+2z\right)}+\sqrt{y\left(z+2x\right)}+z\sqrt{\left(x+2y\right)}\le\sqrt{9}=3\)
\(\Leftrightarrow A\ge\dfrac{9}{3}=3."="\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
quy đồng, nhân chéo ta được (x + 1)(y + 1)(z + 1) \(\ge\)64xyz.
ta có x + 1 = x + x + y + z \(\ge4\sqrt[4]{x^2yz}\)
tương tự với hai ngoặc còn lại và nhân các BĐT đó lại ta được đccm.
Ta có
\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\)
áp dụng bất đẳng thức CS ta có
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\) ;
\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{9}{xy+yz+xz}\)
ta có đánh giá : \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)
\(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{27}\Rightarrow\dfrac{1}{xyz}\ge27\)
\(\Rightarrow1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\ge1+9+27+27=64\)
\(\Rightarrowđpcm\)