cho x,y,z>0 thỏa mãn x+y+z+xy+yz+xz=6.Tìm GTLN của x.y.z
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![](https://rs.olm.vn/images/avt/0.png?1311)
Cho x,y,z >0 thỏa mãn x+y+z = 2. Tìm GTLN của biểu thức
\(P=\sqrt{2x+yz}+\sqrt{2y+xz}+\sqrt{2z+xy}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)
Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)
Cộng vế:
\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)
\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)
P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)
\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)
Đã biết x2 + y2 + z2 \(\ge\)xy + yz + zx
=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)
Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)
= 4
Dấu "=" xảy ra <=> x = 2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{2x+yz}=\sqrt{\left(x+y+z\right)x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+2y+z}{2}\\ \Leftrightarrow P=\sum\sqrt{2x+yz}\le\dfrac{x+2y+z+2x+y+z+x+y+2z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\cdot2=4\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(0\le x,y,z\le1\Rightarrow x^{10}\le x;y^6\le y;z^{2016}\le z;0\le xyz\le1\)
CÓ: \(\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)
=>\(1-xyz+\left(xy+yz+zx\right)-\left(x+y+z\right)\ge0\)
=>\(x+y+z-xy-yz-zx-xyz\le1\)
=>\(x^{10}+y^6+z^{2016}-xy-yz-zx\le1\)
Dấy "=" xảy ra <=> trong 3 số x,y,z có 1 số bằng 0, 2 số bằng 1 hoặc 1 số bằng 1, 2 số bằng 0
Cho x,y,z là các số dương thỏa mãn x+y+z=1. Tìm GTLN của P = \(\sqrt{x+yz}+\sqrt{y+xz}+\sqrt{z+xy}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+xz}+\sqrt{z\left(x+y+z\right)+xy}\)
\(P=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)
\(P\le\dfrac{1}{2}\left(x+y+x+z\right)+\dfrac{1}{2}\left(x+y+y+z\right)+\dfrac{1}{2}\left(x+z+y+z\right)\)
\(P\le2\left(x+y+z\right)=2\)
\(P_{max}=2\) khi \(x=y=z=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ap dung bdt co si ta co:
\(xy+yz+xz>=3\sqrt[3]{\left(xyz\right)^2}\)
=>\(100>3\sqrt[3]{x^2y^2z^2}\)
=>\(\frac{100}{3}>=\sqrt[3]{\left(xyz\right)^2}\)
=>\(\sqrt{\frac{100^3}{3^3}}>=xyz\)
=>\(\frac{1000}{3\sqrt{3}}>=xyz\)
=>\(Amax=\frac{1000}{3\sqrt{3}}\)
xay ra dau bang khi va chi khi x=y=z\(\frac{10}{\sqrt{3}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy+yz+zx\le\dfrac{1}{3}\left(x+y+z\right)^2=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cô si cho gt