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Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).
\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).
Ta có:
\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)
\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).
Ta có:
\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).
\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).
\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).
\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).
\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).
Chứng minh tương tự, ta được:
\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).
Chứng minh tương tự, ta được:
\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).
\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).
\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)
\(\left(4\right)\).
Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).
\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)
(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).
\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(P\ge\frac{\sqrt{5}}{3}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).
Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).
\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Đề Sai sửa lại nha \(A=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{10.\sqrt{z}}{\sqrt{xz}+10\sqrt{x}+10}\)
\(B=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}\)
\(C=\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}\)
\(D=\frac{10.\sqrt{z}}{\sqrt{xz}+10\sqrt{x}+10}\)
\(\Rightarrow C=\frac{\sqrt{x}.\sqrt{y}}{\sqrt{x}.\left(\sqrt{yz}+\sqrt{y}+1\right)}=\frac{\sqrt{xy}}{\sqrt{yzx}+\sqrt{yx}+\sqrt{x}}=\frac{\sqrt{xy}}{10+\sqrt{yx}+\sqrt{x}}\)
(do xyz=100 nên căn xyz=10)
\(\Rightarrow D=\frac{\left(\frac{10.\sqrt{z}}{\sqrt{z}}\right)}{\left(\frac{\sqrt{xz}+10\sqrt{x}+10}{\sqrt{z}}\right)}=\frac{10}{\sqrt{x}+10+\frac{\sqrt{xyz}}{\sqrt{z}}}=\frac{10}{\sqrt{x}+10+\sqrt{xy}}\)(10= căn xyz do xyz=100)
\(\Leftrightarrow A=B+C+D=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\frac{\sqrt{xy}}{10+\sqrt{yx}+\sqrt{x}}+\frac{10}{\sqrt{x}+10+\sqrt{xy}}\)
\(=\frac{\sqrt{xy}+\sqrt{x}+10}{\sqrt{xy}+\sqrt{x}+10}=1\)
T i c k cho mình nha cảm ơn
Ta có x.y.z=100
Suy ra \(\sqrt{xyz}=10\)
Thay \(10=\sqrt{xyz}\) vào A ta được
\(A=\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{xz}+\sqrt{xyz}\sqrt{z}+\sqrt{xyz}}\)
\(A=\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{y}+1+\sqrt{yz}\right)}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{z}}{\sqrt{zx}\left(1+\sqrt{yz}+\sqrt{y}\right)}\)
\(A=\frac{1}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{\sqrt{yz}}{10\left(\sqrt{yz}+\sqrt{y}+1\right)}\)
Mình giải tới đây bí mất rồi ai biết thì làm tiếp rồi chỉ bạn đó nhé
chia cả 2 vế của giả thiết cho xyz rồi đặt 1/x ; 1/y ; 1/z => a ; b ; c
đến đây thì tự làm tiếp đi
Ta có: \(P=\frac{\sqrt{x}}{1+x+xy}+\frac{\sqrt{y}}{1+y+yz}+\frac{\sqrt{z}}{1+z+xz}\)
\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{x+xy+xyz}+\frac{xy\sqrt{z}}{xy+xyz+x^2yz}\)
\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{xy+x+1}+\frac{\sqrt{xy}.\sqrt{xyz}}{xy+x+1}\)
\(P=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{xy}}{xy+x+1}\le\frac{\frac{x+1}{2}+\frac{x\left(y+1\right)}{2}+\frac{xy+1}{2}}{xy+x+1}\) (bđt cosi)
=> \(P\le\frac{x+1+xy+x+xy+1}{2\left(xy+x+1\right)}=\frac{2\left(xy+x+1\right)}{2\left(xy+x+1\right)}=1\)
Dấu "=" xảy ra<=> x = y = z = 1
Vậy MaxP = 1 <=> x = y = z = 1
Theo giả thiết \(\sqrt{\frac{yz}{x}}+\sqrt{\frac{xz}{y}}+\sqrt{\frac{xy}{z}}=3\)
\(\Rightarrow\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}+2x+2y+2z=9\)
Mặt khác , ta có BĐT phụ : \(\frac{yz}{x}+\frac{xz}{y}+\frac{xy}{z}\ge x+y+z\)
\(\Rightarrow9\ge3\left(x+y+z\right)\)
\(\Leftrightarrow x+y+z\le3\)
Áp dụng BĐT Cauchy Shwarz \(\Rightarrow\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)\le9\)
\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le3\)
Ta có : \(P=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{2016}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{2007}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\ge2.\sqrt{9}+\frac{2007}{3}=675\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Chúc bạn học tốt !!!