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Ta có:
\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A được:
\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)(do xy+yz+xz=1)
=>Đpcm
Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.
a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)
b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)
\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)
Thay \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\) ta có
\(1+x=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Tương tự \(1+y=\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\) và \(1+z=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)\)
\(\Rightarrow\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
và \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{z}+\sqrt{x}\right)+\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
Do đó P = 2
\(\frac{x+\left(\sqrt{x}-\sqrt{z}\right)^2}{y+\left(\sqrt{y}-\sqrt{z}\right)^2}=\frac{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-y+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(\sqrt{x}+\sqrt{y}-\sqrt{z}\right)^2-x+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}+2\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)+\left(\sqrt{x}-\sqrt{z}\right)^2}{\left(2\sqrt{x}+\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{y}-\sqrt{z}\right)^2}\)
\(=\frac{\left(\sqrt{x}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(2\sqrt{x}+2\sqrt{y}-2\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}-\sqrt{z}}{\sqrt{y}-\sqrt{z}}\)
TA CÓ:
\(B=\frac{1}{\sqrt{x\left(y+2z\right)}}+\frac{1}{\sqrt{y\left(z+2x\right)}}+\frac{1}{\sqrt{z\left(x+2y\right)}}\ge\frac{1}{\frac{x+y+2z}{2}}+\frac{1}{\frac{y+z+2x}{2}}+\frac{1}{\frac{z+x+2y}{2}}\)
\(\ge\frac{\left(1+1+1\right)^2}{\frac{3}{2}\left(x+y+z\right)}=\frac{18}{3\sqrt{3}}=\frac{6}{\sqrt{3}}\)
DẤU BẰNG XẢY RA:\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
\(\frac{B}{\sqrt{3}}=\frac{1}{\sqrt{3x\left(y+2z\right)}}+\frac{1}{\sqrt{3y\left(z+2x\right)}}+\frac{1}{\sqrt{3z\left(x+2y\right)}}\)
\(\ge\frac{1}{\frac{3x+y+2z}{2}}+\frac{1}{\frac{3y+z+2x}{2}}+\frac{1}{\frac{3z+x+2y}{2}}\ge\frac{2\left(1+1+1\right)^2}{6\left(x+y+z\right)}=\frac{18}{6\sqrt{3}}\)
\(\Rightarrow B\ge\frac{18\sqrt{3}}{6\sqrt{3}}=3\)
Dấu "=" khi \(x=y=z=\frac{1}{\sqrt{3}}\)
Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
vì x2+y2+z2=1 mà x2+y2+z2>=xy+yz+xz suy ra 1>= xy+yz+xz
x2+y2+z2=1 suy ra (x-y)2=1-2xy-z2 ,(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]=\)
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)(do (x-y)2=1-2xy-z2(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2)
theo bdt cosi ta có:
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2z\sqrt{2xy}+2y\sqrt{2xz}+2x\sqrt{2yz}\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-3\sqrt[3]{\left(2z\sqrt{2xy}.2y\sqrt{2xz}.2x\sqrt{2yz}\right)}\)
\(=\sqrt{3}+\frac{\sqrt{3}}{2}[1-2\sqrt{2}.\sqrt[3]{xyz^2}]\)\(=\sqrt{3}\left(1+\frac{1}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)=\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra
\(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\left(doxy+yz+xz\le1\right)\)
ta giả sử:
\(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\Leftrightarrow\sqrt{3}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\) mà \(\sqrt{3}>\frac{3}{2}\)
suy ra \(\frac{3}{2}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\)(luôn đúng) suy ra điều giả sử trên là đúng
hay \(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
mà \(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\),\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)\(\le\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)(đpcm)
em mới có lớp 8, nếu em làm sai cho em xin lỗi nha anh
chứng minh $\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}\leq \frac{3}{2}\sqrt{(x+1)(y+1)(z+1)}$ - Bất đẳng thức và cực trị - Diễn đàn Toán học
Sử dụng Bunhiacopxki:
\(\sqrt{x\left(1-x\right)}+\sqrt{y\left(1-y\right)}+\sqrt{z\left(1-z\right)}\le\sqrt{\left(x+y+z\right)\left(1-x+1-y+1-z\right)}=\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)