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\(x+\sqrt{x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{x^2+yz+x\left(z+y\right)}\)
\(\ge x+\sqrt{2\sqrt{x^2yz}+x\left(y+z\right)}=x+\sqrt{x\cdot2\sqrt{yz}+x\left(y+z\right)}=x+\sqrt{x\left(y+z+2\sqrt{yz}\right)}\)
\(=x+\sqrt{x\left(\sqrt{y}+\sqrt{z}\right)^2}=x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow\frac{x}{x+\sqrt{x+yz}}\le\frac{x}{x+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
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
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 \(\Leftrightarrow x=y=z=\frac{1}{3}\)
Đặt \(\left(x;2y;3z\right)=\left(a;b;c\right)\Rightarrow a+b+c=2\)
\(S=\sqrt{\dfrac{ab}{ab+2c}}+\sqrt{\dfrac{bc}{bc+2a}}+\sqrt{\dfrac{ca}{ca+2b}}\)
\(S=\sqrt{\dfrac{ab}{ab+c\left(a+b+c\right)}}+\sqrt{\dfrac{bc}{bc+a\left(a+b+c\right)}}+\sqrt{\dfrac{ca}{ca+b\left(a+b+c\right)}}\)
\(S=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ca}{\left(a+b\right)\left(b+c\right)}}\)
\(S\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\Rightarrow x;y;z\)
\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)
áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương
ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)
ta có :
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)
lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)
ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :
\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)
\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)
vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673
\(xy+yz+zx\le3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)
\(P=\frac{1}{\sqrt{x^2+y^2+x^2+xy}}+\frac{1}{\sqrt{y^2+z^2+y^2+yz}}+\frac{1}{\sqrt{z^2+x^2+z^2+zx}}\)
\(P\le\frac{1}{\sqrt{x^2+3xy}}+\frac{1}{\sqrt{y^2+3yz}}+\frac{1}{\sqrt{z^2+3zx}}=\frac{4}{2\sqrt{4x\left(x+3y\right)}}+\frac{4}{2\sqrt{4y\left(y+3z\right)}}+\frac{1}{2\sqrt{4z\left(z+3x\right)}}\)
\(P\le4\left(\frac{1}{4x+x+3y}+\frac{1}{4y+y+3z}+\frac{1}{4z+z+3x}\right)=4\left(\frac{1}{5x+3y}+\frac{1}{5y+3z}+\frac{1}{5z+3x}\right)\)
\(P\le\frac{4}{64}\left(\frac{5}{x}+\frac{3}{y}+\frac{5}{y}+\frac{3}{z}+\frac{5}{z}+\frac{3}{x}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{3}{2}\)
\(P_{max}=\frac{3}{2}\) khi \(x=y=z=1\)
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
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
Đặt \(\hept{\begin{cases}x=a\\2y=b\\3z=c\end{cases}}\left(a;b;c>0\right)\Rightarrow a+b+c=2\)
Khi đó \(S=\Sigma\sqrt{\frac{\frac{ab}{2}}{\frac{ab}{2}+c}}=\Sigma\sqrt{\frac{ab}{ab+2c}}=\Sigma\sqrt{\frac{ab}{ab+\left(a+b+c\right)c}}\)
\(=\Sigma\sqrt{\frac{ab}{ab+bc+ca+c^2}}=\Sigma\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}\)
Áp dụng bđt Cô-si có
\(S\le\frac{\Sigma\left(\frac{a}{a+c}+\frac{b}{b+c}\right)}{2}=\frac{3}{2}\)
