cho x,y,z thỏa mãng: x^2=yz, y^2=xz,z^2=xy. cmrx=y=z
cho x,y,z thỏa mãng: x^2=yz, y^2=xz,z^2=xy. cmrx=y=z
cho x,y,z khác 0 thỏa mãn xy/x+y=yz/y+z=xz/x+z
tính giá trị của M=\(\frac{x^2+y^2+z^2}{xy+xz+yz}\)
Cho 3 số dương x,y,z thỏa mãn: xy + yz + xz = 671
\(CM:\dfrac{x}{x^2-yz+2013}+\dfrac{y}{y^2-xz+2013}+\dfrac{z}{z^2-xy+2013}\ge\dfrac{1}{x+y+z}\)
Ta có:
\(VT=\dfrac{x^2}{x^3-xyz-2013x}+\dfrac{y^2}{y^3-xyz-2013y}+\dfrac{z^2}{z^3-xyz-2013z}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013.\left(z+y+z\right)}\)
\(VT=\dfrac{\left(x+y+x\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right).\left(xy+yz+xz\right)-xyz\right]}\)
\(VT=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}\)
\(VT=\dfrac{1}{x+y+z}=VP\)
\(\Rightarrow\) Đpcm.
cho 3 số x,y,z thỏa mãn x^2+y^2 +z^2=xy+yz+xz và x+y+z=-3 .Tính B = x^2020 +y^2021+z^2022
\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)
Cho x, y, z >0 thỏa mãn x + y + z= xyz
CMR: \(\dfrac{x}{x^2+yz}+\dfrac{y}{y^2+xz}+\dfrac{z}{z^2+xy}\le\dfrac{\sqrt{3}}{2}\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
\(VT\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}\)
\(VT\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\le\dfrac{1}{2}\sqrt{3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}=\dfrac{\sqrt{3}}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
cho x,y,z thỏa mãn \(x+y+z\le\dfrac{3}{2}\) . tìm GTNN của \(P=\dfrac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\dfrac{y\left(xz+1\right)^2}{y^2\left(xy+1\right)}+\dfrac{z\left(xy+1\right)^2}{x^2\left(yz+1\right)}\)
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.
(\sqrt((x+yz)(y+xz)))/(xy+z)+(\sqrt((y+xz)(z+xy)))/(x+yz)+(\sqrt((x+yz)(z+xy)))/(y+xz)
Với x,y,z>0 thỏa mãn x+y+z=1
cho 3 số x,y,z>0 thỏa mãn x^2+y^2+z^2=3.tìm Min xy/z+yz/x+xz/y
cho x,y,z là các số thực thỏa mãn x2+y2+z2=1. tìm GTLN của bt M=2(xy+yz+xz)+(xy-xz)2+(yz-xy)2+(xz-yz)2