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\(P=\frac{1}{x^2+y^2+z^2}+\frac{2009}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}+\frac{2007}{xy+yz+zx}\)
\(P\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{2007}{\frac{1}{3}\left(x+y+z\right)^2}\)
\(P\ge\frac{9}{\left(x+y+z\right)^2}+\frac{6021}{\left(x+y+z\right)^2}=\frac{6030}{\left(x+y+z\right)^2}\ge\frac{6030}{3^2}=670\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT Côsi dưới dạng engel, ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\)
⇒\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right).\frac{9}{x+y+z}\) = 9
Dấu "=" xảy ra ⇔ x = y = z
\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
Theo bài ra ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\Rightarrow x+y+z=xyz\)
Do:\(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự: \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\);
\(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(z+y\right)\left(x+y\right)}\)
\(A=\sqrt{\frac{x^2}{yz\left(1+x^2\right)}}+\sqrt{\frac{y^2}{zx\left(1+y^2\right)}}+\sqrt{\frac{z^2}{xy\left(1+z^2\right)}}\)
\(A=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng bất đẳng thức Cô si \(\frac{a+b}{2}\ge\sqrt{ab}\), dấu "=" xảy ra khi \(a=b\)
Ta có \(\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\);
\(\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\);
\(\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\le\frac{1}{2}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\)
\(A\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{y+x}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{2}\)
Vậy \(A\le\frac{3}{2}\). Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
M giải thích cho t chỗ sao mà \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\) đc vậy?
Với cả từ dòng này xuống dòng này nữa.
Sao mà tin đc dấu " = " xảy ra khi nào vậy?
\(\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Vì \(x,y,z\) khác nhau nên \(x+y+z=0\)
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(P=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\frac{\left(-x\right)\cdot\left(-y\right)\cdot\left(-z\right)}{xyz}=-1\)
Vậy...
Hướng dẫn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
Thay vào:\(x^2+2yz=x^2+yz+yz=x^2+yz-xy-zx=x\left(x-y\right)-z\left(x-y\right)=\left(x-y\right)\left(x-z\right)\)
Tương tự thay vào mà quy đồng
a) \(A=\frac{x\left(x^2-yz\right)}{x+y+z}+\frac{y\left(y^2-zx\right)}{x+y+z}+\frac{z\left(z^2-xy\right)}{x+y+z}\)
\(=\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)
\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)}{x+y+z}\)
\(=x^2+y^2+z^2-xy-yz-xz\)
b) \(B=\frac{2}{3}.\left[\frac{3}{4x^2+4x+4}+\frac{3}{4x^2-4x+4}\right]\)
\(=\frac{2}{3}.\frac{3}{4}.\left(\frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right)\)
\(=\frac{1}{2}.\frac{x^2-x+1+x^2+x+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{1}{2}.\frac{2\left(x^2+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(=\frac{x^2+1}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
(vì \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
và \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\))