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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
BĐT <=> \(\sqrt{\frac{x+yz}{xyz}}+\sqrt{\frac{y+xz}{xyz}}+\sqrt{\frac{z+xy}{xyz}}\ge1+\sqrt{\frac{1}{xy}}+\sqrt{\frac{1}{yz}}+\sqrt{\frac{1}{xz}}\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
Khi đó \(a+b+c=1\)
BĐT <=>\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
Ta có \(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)
Khi đó \(VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=VP\)(ĐPCM)
Dấu bằng xảy ra khi x=y=z=3
BĐT cho tương đương với
\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Với \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z};a+b+c=1\)
Ta có:
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}\)
\(=\sqrt{a^2+a\left(b+c\right)+bc}\ge\sqrt{a^2+2a\sqrt{bc}+bc}=a+\sqrt{bc}\)
Tương tự
\(\sqrt{b+ca}\ge b+\sqrt{ca};\sqrt{c+ab}\ge c+\sqrt{ab}\)
Từ đó ta có đpcm
Dấu "=" xảy ra khi x=y=z=3
\(\sqrt{x\left(1-y\right)\left(1-z\right)}=\sqrt{x\left(yz-y-z+1\right)}=\sqrt{x\left(yz-y-z+x+y+z+2\sqrt{xyz}\right)}\)
\(=\sqrt{x\left(yz+x+2\sqrt{xyz}\right)}=\sqrt{x^2+2x\sqrt{xyz}+xyz}=\sqrt{\left(x+\sqrt{xyz}\right)^2}\)
\(=x+\sqrt{xyz}\)
Tương tự: \(\sqrt{y\left(1-x\right)\left(1-z\right)}=y+\sqrt{xyz}\) ; \(\sqrt{z\left(1-x\right)\left(1-y\right)}=z+\sqrt{xyz}\)
\(\Rightarrow VT=x+y+z+3\sqrt{xyz}=1-2\sqrt{xyz}+3\sqrt{xyz}=1+\sqrt{xyz}\) (đpcm)
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(x^2+y^2+z^2\)
\(=\frac{x^2+y^2}{2}+\frac{y^2+z^2}{2}+\frac{z^2+x^2}{2}\)
\(\ge xy+yz+zx\)
\(=\frac{xy+yz}{2}+\frac{yz+zx}{2}+\frac{zx+xy}{2}\)
\(\ge\frac{2\sqrt{xy^2z}}{2}+\frac{2\sqrt{xyz^2}}{2}+\frac{2\sqrt{x^2yz}}{2}\)
\(=\sqrt{xyz}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\frac{1}{x^3\left(y+z\right)}+\frac{1}{y^3\left(z+x\right)}+\frac{1}{z^3\left(x+y\right)}\)
\(=\frac{y^2z^2}{x\left(y+z\right)}+\frac{z^2x^2}{y\left(z+x\right)}+\frac{x^2y^2}{z\left(x+y\right)}\)
\(\ge\frac{\left(xy+yz+zx\right)^2}{2\left(xy+yz+zx\right)}=\frac{xy+yz+zx}{2}\ge\frac{3\sqrt[3]{x^2y^2z^2}}{2}=\frac{3}{2}\)
Bạn thiếu giả thiết \(x,y,z>0\) nhé.
Theo giả thiết \(xyz=xy+yz+zx.\) Từ đó ta có\(\sqrt{x+yz}=\sqrt{\frac{x^2+xyz}{x}}=\sqrt{\frac{x^2+xy+yz+zx}{x}}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}.\)
Theo bất đẳng thức Bunhiacốpxki, \(\sqrt{\left(x+y\right)\left(x+z\right)}\ge x+\sqrt{yz}.\) Do đó
\(\sqrt{x+yz}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x}}\ge\frac{x+\sqrt{yz}}{\sqrt{x}}=\sqrt{x}+\sqrt{\frac{yz}{x}}\), hay ta có \(\sqrt{x+yz}\ge\sqrt{x}+\sqrt{\frac{yz}{x}}.\)
Tương tự ta có hai bất đẳng thức nữa\(\sqrt{y+zx}\ge\sqrt{y}+\sqrt{\frac{xz}{y}},\sqrt{z+xy}\ge\sqrt{z}+\sqrt{\frac{xy}{z}}\). Cộng cả ba bất đẳng thức lại cho ta
\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{\frac{yz}{x}}+\sqrt{y}+\sqrt{\frac{zx}{y}}+\sqrt{z}+\sqrt{\frac{xy}{z}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\left(\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+\sqrt{\frac{xy}{z}}\right)\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+zx}{\sqrt{xyz}}\)
\(\Leftrightarrow\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}.\) (ĐPCM)
Đề bài sai với $x=0; y=z=1$