Cho \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\left(x,y,z\ne0\right)\). Tính \(\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}\)
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Áp dụng bất đẳng thức AM - GM:
\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).
Áp dụng bất đẳng thức AM - GM ta có:
\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).
Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).
Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)
\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).
Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)
Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)
\(\Rightarrow P\ge\dfrac{15}{2}\).
Vậy...
Áp dụng bất đẳng thức AM - GM:
P≥33√(xy+1)(yz+1)(zx+1)xyz.
Áp dụng bất đẳng thức AM - GM ta có:
xy+1=xy+14+14+14+14≥55√xy44.
Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.
Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412
⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.
Mà xyz≤(x+y+z)327=18
Nên (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258
⇒P≥152.
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)
Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)
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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)
\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)
\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)
\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)
Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)
Suy ra:
\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)
\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\)
\(y^2-yz+z^2=y^2+\left(z-y\right)y\le y^2\Rightarrow\dfrac{1}{y^2-yz+z^2}\ge\dfrac{1}{y^2}\)
Tương tự: \(\dfrac{1}{z^2-xz+x^2}\ge\dfrac{1}{x^2}\)
\(\Rightarrow P\ge\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{x^2-xy+y^2}+\dfrac{x^2-xy+y^2}{x^2y^2}+\dfrac{1}{xy}\)
\(P\ge2\sqrt{\dfrac{x^2-xy+y^2}{x^2y^2\left(x^2-xy+y^2\right)}}+\dfrac{1}{xy}=\dfrac{3}{xy}\ge\dfrac{12}{\left(x+y\right)^2}\ge\dfrac{12}{\left(x+y+z\right)^2}=3\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\) và hoán vị
![](https://rs.olm.vn/images/avt/0.png?1311)
Với a; b dương, nếu \(a\ge b\) thì \(\dfrac{1}{a}\le\dfrac{1}{b}\)
Áp dụng BĐT Cô-si cho mẫu số vế trái ta được:
\(\dfrac{1}{x^2+yz}+\dfrac{1}{y^2+xz}+\dfrac{1}{z^2+xy}\le\dfrac{1}{2x\sqrt{yz}}+\dfrac{1}{2y\sqrt{xz}}+\dfrac{1}{2z\sqrt{xy}}\)
\(\Rightarrow VT\le\dfrac{\sqrt{yz}}{2xyz}+\dfrac{\sqrt{xz}}{2xyz}+\dfrac{\sqrt{xy}}{2xyz}=\dfrac{\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}{2xyz}\)
Tiếp tục dùng Cô-si cho tử số:
\(VT\le\dfrac{\dfrac{y+z}{2}+\dfrac{x+z}{2}+\dfrac{x+y}{2}}{2xyz}=\dfrac{x+y+z}{2xyz}\)
\(\Rightarrow VT\le\dfrac{1}{2}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\) (đpcm)
Dấu "=" xảy ra khi x=y=z
![](https://rs.olm.vn/images/avt/0.png?1311)
thay 1=x+y+z vào nhá , ví dụ x=x(x+y+z) rồi phân tích đa thức thành nhân tử!
thay 1=x+y+z vào nhá , ví dụ x=x(x+y+z) rồi phân tích đa thức thành nhân tử!
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\dfrac{x^2+x-6}{x^3-4x^2-18x+9}=\dfrac{x^2+3x-2x-6}{x^3+3x^2-7x^2-21x+3x+9}\)
\(=\dfrac{x\left(x+3\right)-2\left(x+3\right)}{x^2\left(x+3\right)-7x\left(x+3\right)+3\left(x+3\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x^2-7x+3\right)\left(x+3\right)}=\dfrac{x-2}{x^2-7x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x,y,z\ne0\)
-Ta c/m: -Với \(a+b+c=0\) thì: \(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0.\left(a^2+b^2+c^2-ab-bc-ca\right)=0\left(đpcm\right)\)
-Quay lại bài toán:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo bài ra :
\(\dfrac{x^2-yz}{x\left(1-yz\right)}=\dfrac{y^2-xz}{y\left(1-xz\right)}\)
\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)
\(\Leftrightarrow x^2y-x^3yz-y^2z+xx^2z^2=xy^2-xy^3z-x^2z+x^2yz^2=0\)
\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)+xyz^2\left(x-x\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right)\right]=0\)
\(\Leftrightarrow\left(x-y\right)\left(xy+xz+yz-xyz\left(x+y+z\right)\right)=0\)
Mà theo đề bài :
\(x\ne y\Rightarrow xy+xz+yz-xyz\left(x+y+z\right)=0\)
\(\Leftrightarrow xy+xz+yz=xyz\left(x+y+z\right)\)
\(\Leftrightarrow\dfrac{xy}{xyz}+\dfrac{xz}{xyz}+\dfrac{yz}{xyz}=\dfrac{xyz\left(z+y+x\right)}{xyz}\)
\(\Leftrightarrow\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}=x+y+z\left(đpcm\right)\)
Ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=-\dfrac{1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}+3\dfrac{1}{x}\dfrac{1}{y}.\left(-\dfrac{1}{z}\right)=0\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)
\(\Leftrightarrow xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{3}{xyz}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
Vậy...