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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
Biến đổi tương đương, dễ dàng chứng minh Bđt:
\(\frac{4}{\left(x+y\right)^2}+\frac{4}{\left(x+z\right)^2}\ge\frac{4}{x^2+yz}\)\(\Rightarrow VT\ge\frac{x^2}{yz}+\frac{4}{x^2+yz}\)
Từ \(3y^2z^2+x^2=2\left(x+yz\right)\) ta có:
\(3y^2z^2+x^2\le x^2+1+2yz\)
\(\Rightarrow3y^2z^2-2yz-1\le0\Rightarrow yz\le1\)
Khi đó:
\(VT\ge x^2+\frac{4}{x^2+1}=\left(x^2+1\right)+\frac{4}{x^2+1}-1\ge3\)
Dấu = khi x=y=z=1
\(H\ge\frac{\left(x+y\right)^2}{2xy\left(x+y^3\right)}+\frac{\left(y+z\right)^2}{2yz\left(y+z\right)}+\frac{\left(z+x\right)^2}{2zx\left(z+x\right)}=\frac{1}{2xy\left(x+y\right)}+\frac{1}{2yz\left(y+z\right)}+\frac{1}{2zx\left(z+x\right)}\)
\(\Rightarrow H\ge\frac{9}{2}.\frac{1}{xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)}\)
Ta chứng minh BĐT phụ sau:
\(x^3+y^3\ge xy\left(x+y\right)\)
\(\Leftrightarrow x^3-x^2y+y^3-xy^2\ge0\)
\(\Leftrightarrow x^2\left(x-y\right)-y^2\left(x-y\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (luôn đúng)
Vậy BĐT phụ được chứng minh
Hoàn toàn tương tự: \(y^3+z^3\ge yz\left(y+z\right)\); \(z^3+x^3\ge zx\left(z+x\right)\)
\(\Rightarrow H\ge\frac{9}{2}.\frac{1}{x^3+y^3+y^3+z^3+z^3+x^3}=\frac{9}{4\left(x^3+y^3+z^3\right)}=\frac{9}{32}\)
\(H_{min}=\frac{9}{32}\) khi \(x=y=z=\frac{2\sqrt{3}}{3}\)
Ta co:
\(x+y+z=2\)
\(\Leftrightarrow\left(x+y+z\right)^2=4\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta lai co:
\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(z+x\right)\)
\(1+y^2=xy+yz+zx+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=xy+yz+zx+z^2=\left(y+z\right)\left(z+x\right)\)
Thay vao P ta duoc:
\(P=\Sigma x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(y+z\right)\left(z+x\right)}{\left(x+y\right)\left(z+x\right)}}=\Sigma\left(y+z\right)=2\left(x+y+z\right)=2.2=4\)
à xin lỗi nha cái hạng tử cuối cùng là \(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\) mới đúng