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TA CÓ:
\(Q=\frac{x\left(\sqrt{x+zy}-x\right)}{x+yz-x^2}+\frac{y\left(\sqrt{y+zx}-y\right)}{y+zx-y^2}+\frac{z\left(\sqrt{xy+z}-z\right)}{z+xy-z^2}\)
\(=\frac{x\left(\sqrt{x\left(x+y+z\right)+yz}-x\right)}{x\left(x+y+z\right)+yz-x^2}+\frac{y\left(\sqrt{y\left(x+y+z\right)+zx}-y\right)}{y\left(x+y+z\right)-y^2+zx}+\frac{z\left(\sqrt{xy+z\left(x+y+z\right)}-z\right)}{z\left(x+y+z\right)+xy-z^2}\)
\(=\frac{x\left(\sqrt{\left(x+y\right)\left(z+x\right)}-x\right)}{xy+yz+zx}+\frac{y\left(\sqrt{\left(x+y\right)\left(y+z\right)}-y\right)}{xy+yz+zx}+\frac{z\left(\sqrt{\left(y+z\right)\left(z+x\right)}-z\right)}{xy+yz+za}\)
ÁP DỤNG BĐT CÔ-SI TA ĐƯỢC:
\(Q\le\frac{x\left(\frac{x+y+z+x}{2}-x\right)}{xy+zx+yz}+\frac{y\left(\frac{x+y+z+y}{2}-y\right)}{xy+yz+zx}+\frac{z\left(\frac{x+y+z+z}{2}-z\right)}{xy+yz+zx}\)
\(=\frac{xy+zx}{2\left(xy+yz+zx\right)}+\frac{xy+yz}{2\left(xy+yz+zx\right)}+\frac{yz+zx}{2\left(xy+yz+zx\right)}=1\)
DẤU BẰNG XẢY RA \(\Leftrightarrow x=y=z=\frac{1}{3}\)
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 !!!
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
ta có 3x + yz = x2 + xy + yz + zx = (x+y)(x+z)
do đó:
\(\frac{x}{x+\sqrt{3x+yz}}=\frac{x\left(\sqrt{x^2+xy+yz+zx}-x\right)}{\left(\sqrt{x^2+xy+yz+zx}+x\right)\left(\sqrt{x^2+xy+yz+zx}-x\right)}\)
= \(\frac{x\left(\sqrt{\left(x+y\right)\left(x+z\right)}-x\right)}{xy+yz+zx}\le\frac{x\left(\frac{x+y+x+z}{2}-x\right)}{xy+yz+zx}\)\(\le\frac{x\left(y+z\right)}{2\left(xy+yz+zx\right)}\)
tương tự với 2 số hạng còn lại nên ta được: P\(\le\)1. đpcm
Lời giải:
Do $x+y+z=1$ nên:
$x+yz=x(x+y+z)+yz=(x+y)(x+z)=(x+y)(z+x)\geq (\sqrt{xz}+\sqrt{xy})^2$ theo BĐT Bunhiacopxky
$\Rightarrow \sqrt{x+yz}\geq \sqrt{xz}+\sqrt{xy}$
$\Rightarrow \frac{x}{x+\sqrt{x+yz}}\leq \frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}$
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:
$Q\leq \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1$
Vậy $Q_{\max}=1$ khi $x=y=z=\frac{1}{3}$
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
Cho x,y,z là các số dương thỏa mãn x+y+z=1. Tìm GTLN của P = \(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\)
\(P=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+xz}+\sqrt{z\left(x+y+z\right)+xy}\)
\(P=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)
\(P\le\dfrac{1}{2}\left(x+y+x+z\right)+\dfrac{1}{2}\left(x+y+y+z\right)+\dfrac{1}{2}\left(x+z+y+z\right)\)
\(P\le2\left(x+y+z\right)=2\)
\(P_{max}=2\) khi \(x=y=z=\dfrac{1}{3}\)
\(x+\sqrt{x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{x^2+yz+x\left(z+y\right)}\)
\(\ge x+\sqrt{2\sqrt{x^2yz}+x\left(y+z\right)}=x+\sqrt{x\cdot2\sqrt{yz}+x\left(y+z\right)}=x+\sqrt{x\left(y+z+2\sqrt{yz}\right)}\)
\(=x+\sqrt{x\left(\sqrt{y}+\sqrt{z}\right)^2}=x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow\frac{x}{x+\sqrt{x+yz}}\le\frac{x}{x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
tương tự :
\(\frac{y}{y+\sqrt{y+xz}}\le\frac{\sqrt{y}}{\sqrt{y}+\sqrt{x}+\sqrt{z}}\)
\(\frac{z}{z+\sqrt{z+xy}}\le\frac{\sqrt{z}}{\sqrt{z}+\sqrt{x}+\sqrt{y}}\)
cộng vế theo vế ta được
\(\frac{x}{x+\sqrt{x+yz}}+\frac{y}{y+\sqrt{y+zx}}+\frac{z}{z+\sqrt{z+xy}}\le\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
dấu "=" xảy tra khi x=y=z=1/3
cái này thì chịu