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Từ dữ kiện đề bài => x + y + z = xyz
Ta có :
\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+xyz.x}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)
\(=\frac{\sqrt{x}}{\sqrt{x+z}}.\frac{\sqrt{x}}{\sqrt{x+y}}\le\frac{1}{2}.\left(\frac{x}{x+z}+\frac{x}{x+y}\right)\)
Tương tự với hai hạng tử còn lại , suy ra
\(Q\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)+\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy Max = 3/2 <=> x = y = z
Nguồn : Đinh Đức Hùng
Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)
\(\Rightarrow\)\(x+y+z=xyz\)
Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)
Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)
\(Q=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)
Dấu "=" xảy ra khi A = B :
Ta được :
\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)
Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)với a,b>0
Ta có: \(\frac{4xy}{z+1}=\frac{4xy}{2z+x+y}\le\frac{xy}{x+z}+\frac{xy}{y+z}\)
Tương tự: \(\frac{4yz}{x+1}\le\frac{yz}{x+y}+\frac{yz}{x+z}\)
\(\frac{4zx}{y+1}\le\frac{zx}{y+x}+\frac{zx}{y+z}\)
\(\Rightarrow4\left(\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\right)\le\frac{xy}{x+z}+\frac{xy}{y+z}+\frac{yz}{x+y}+\frac{yz}{x+z}+\frac{zx}{y+x}+\frac{zx}{y+z}=x+y+z=1\)
\(\Rightarrow\frac{xy}{z+1}+\frac{yz}{x+1}+\frac{zx}{y+1}\le\frac{1}{4}\)
Dấu "=" xảy ra khi: x=y=z>0
Bài 2:
+) Với y=0 <=> x=0
Ta có: 1-xy= 12 (đúng)
+) Với \(y\ne0\)
Ta có: \(x^6+xy^5=2x^3y^2\)
\(\Leftrightarrow x^6-2x^3y^2+y^4=y^4-xy^5\)
\(\Leftrightarrow\left(x^3-y^2\right)^2=y^4\left(1-xy\right)\)
\(\Rightarrow1-xy=\left(\frac{x^3-y^2}{y^2}\right)^2\)
Bạn tham khảo tại đây:
Câu hỏi của trieu dang - Toán lớp 8 - Học toán với OnlineMath
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{\left(yz+xz+xy\right)}{xyz}=0\)
\(\Rightarrow yz+zx+xy=0\)
Ta có : \(x^2+2yz=x^2+yz+yz\)
\(=x^2+yz-zx-xy\)
\(=x\left(x-z\right)-y\left(x-z\right)\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự : \(y^2+2xz=y^2+xz+xz\)
\(=y^2+xz-xy-yz\)
\(=y\left(y-x\right)+z\left(x-y\right)\)
\(=\left(x-y\right)\left(z-y\right)\)
\(z^2+2xy=\left(x-z\right)\left(y-z\right)\)
\(\Rightarrow M=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\) \(M=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(M=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)
\(\Leftrightarrow xy=-yz-zx;yz=-xy-zx;zx=-xy-yz\)
Ta có: x2+2yz=x2+yz+yz=x2+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)
Tương tự: \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)
A= \(\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)=\(\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{xy\left(x-y\right)-xz\left(x-y+y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{xy\left(x-y\right)-xz\left(x-y\right)-xz\left(y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)\(=\frac{\left(xy-xz\right)\left(x-y\right)-\left(xz-yz\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{x\left(y-z\right)\left(x-y\right)-z\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=1\)