Bài 10. Cho 3 số dương x,y,z thỏa mãn: (√(xy/z)+√(xz/y)+√(yz/x)) = 3
Tìm GTNN của: P = (√x+√y+√z) + (2016/(√x+√y)) + (2016/√z)
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17 tháng 2 2016
thay 2016=xy+yz+xz vào các mẫu
dùng Cô-Si đảo vào từng phân số
sẽ dễ dàng chứng minh đc :D
TN
1 tháng 2 2017
B1:x^2+2016=xy+yz+xz+x^2=...
tuong tu
y^2+2016=... ; z^2+2016=....
B2:bdt am-gm
S
21 tháng 2 2020
\(0\le x,y,z\le1\Rightarrow x^{10}\le x;y^6\le y;z^{2016}\le z;0\le xyz\le1\)
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
KK
4 tháng 4 2017
\(VT=\sqrt{\dfrac{yz}{x^2+xy+yz+xz}}+\sqrt{\dfrac{xy}{y^2+xy+yz+xz}}+\sqrt{\dfrac{xz}{z^2+xy+yz+xz}}\)
\(VT=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\\\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{y}{y+z}}{2}\\\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{z}{y+z}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{x+z}+\dfrac{x}{x+z}\right)}{2}\)
\(\Rightarrow VT\le\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{y^2+2016}}+\sqrt{\dfrac{xz}{z^2+2016}}\le\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(x=y=z=4\sqrt{42}\)
4 tháng 4 2017
Sửa đề:\(\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}\)
Giải
Ta có:
\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{z^2+xy+xz+yz}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\sqrt{\dfrac{yz}{x^2+2016}}\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right);\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{z}{y+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(\Sigma\sqrt{\dfrac{xy}{z^2+2016}}\le\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=4\sqrt{42}\)
D
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
\(xy+yz+zx\le\dfrac{1}{3}\left(x+y+z\right)^2=\dfrac{4}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
KK
21 tháng 1 2017
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
25 tháng 4 2021
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)