cho các số thực dương x,y,z thoả mãn \(\sqrt{x}\) + \(\sqrt{y}\) + \(\sqrt{z}\) = 1
chứng minh rằng : \(\sqrt{\dfrac{xy}{x+y+2z}}\) + \(\sqrt{\dfrac{yz}{y+z+2x}}\) + \(\sqrt{\dfrac{zx}{z+x+2y}}\) ≤ \(\dfrac{1}{2}\)
cho các số thực dương thoả mãn: \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\)
CMR: \(\sqrt{\dfrac{xy}{x+y+2z}}+\sqrt{\dfrac{yz}{y+z+2x}}\sqrt{\dfrac{zx}{z+x+zy}}\le\dfrac{1}{2}\)
Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)
Tương tự cũng có:
\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)
\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
Cộng vế với vế ta được:
\(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)
\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)
chi các số thực dương x,y,z thỏa mãn \(x^4+y^4+z^4=3\)
Tìm GTNN của T=\(\sqrt{\dfrac{yz}{7-2x}}+\sqrt{\dfrac{zx}{7-2y}}+\sqrt{\dfrac{xy}{7-2z}}\)
với các số thực dương x,y,z và xyz=1
chứng minh đẳng thức
\(\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge3\sqrt{3}\)
\(A=\dfrac{\sqrt{x^3+y^3+1}}{xy}+\dfrac{\sqrt{y^3+z^3+1}}{yz}+\dfrac{\sqrt{z^3+x^3+1}}{zx}\)
\(\dfrac{\sqrt{x^3+y^3+1}}{xy}=\dfrac{\sqrt{x^3+y^3+xyz}}{xy}\ge\dfrac{\sqrt{xy\left(x+y\right)+xyz}}{xy}=\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}\ge\dfrac{\sqrt{xy.3^3\sqrt{xyz}}}{xy}=\dfrac{\sqrt{3xy}}{xy}=\dfrac{\sqrt{3}}{\sqrt{xy}}\)
\(\dfrac{\sqrt{y^3+z^3+1}}{yz}\ge\dfrac{\sqrt{3}}{\sqrt{yz}}\)
\(\dfrac{\sqrt{z^3+x^3+1}}{zx}\ge\dfrac{\sqrt{3}}{\sqrt{zx}}\)
\(\Rightarrow A\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.xz}}}=3\sqrt{3}.\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)
Cho x, y, z là ba số thực dương. Tìm giá trị nhỏ nhất của biểu thức:
S = \(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}+\dfrac{\sqrt{y^2-yz+z^2}}{2x+y+z}+\dfrac{\sqrt{z^2-zx+x^2}}{x+2y+z}\)
Ta có x2-xy+y2=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)
=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...
Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :
S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\) (Netbit)
Cho các số thực dương x,y,z thoả mãn \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\)
Chứng minh : \(\sqrt{\dfrac{xy}{x+y+2z}}+\sqrt{\dfrac{yz}{y+z+2x}}+\sqrt{\dfrac{zx}{Z+x+2y}}\le\dfrac{1}{2}\)
Đặt \(\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)\rightarrow\left(a;b;c\right)\)\(\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c>0\end{matrix}\right.\)
Và \(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}+\dfrac{bc}{\sqrt{b^2+c^2+2a^2}}+\dfrac{ca}{\sqrt{c^2+a^2+2b^2}}\le\dfrac{1}{2}\)
Ta có:\(\dfrac{ab}{\sqrt{a^2+b^2+2c^2}}=\dfrac{2ab}{\sqrt{\left(1+1+2\right)\left(a^2+b^2+2c^2\right)}}\)
\(\le\dfrac{2ab}{a+b+2c}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{bc+ac}{a+b}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=\dfrac{1}{9}\)
cho các số thực dưong x,y,z thỏa mãn : x2+y2+z2=3
chứng minh rằng : \(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{zx}}+\dfrac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
nhờ mn giúp mk bài này vs ạ
mk đang cần gấp !
cảm ơn mn nhiều
Đặt \(\left(\sqrt[3]{x};\sqrt[3]{y};\sqrt[3]{z}\right)=\left(a;b;c\right)\) \(\Rightarrow a^6+b^6+c^6=3\)
\(a^6+a^6+a^6+a^6+a^6+1\ge6a^5\)
Tương tự: \(5b^6+1\ge6b^5\) ; \(5c^6+1\ge6c^5\)
Cộng vế với vế: \(18=5\left(a^6+b^6+c^6\right)+3\ge6\left(a^5+b^5+c^5\right)\)
\(\Rightarrow3\ge a^5+b^6+b^5\)
BĐT cần chứng minh: \(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge a^3b^3+b^3c^3+c^3a^3\)
Ta có:
\(\dfrac{a^3}{bc}+\dfrac{b^3}{ca}+\dfrac{c^3}{ab}\ge\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge a+b+c\) (1)
Mà \(3\left(a+b+c\right)\ge\left(a^5+b^5+c^5\right)\left(a+b+c\right)\ge\left(a^3+b^3+c^3\right)^2\ge3\left(a^3b^3+b^3c^3+c^3a^3\right)\)
\(\Rightarrow a+b+c\ge a^3b^3+b^3c^3+c^3a^3\) (2)
Từ (1);(2) \(\Rightarrow\) đpcm
Cho các số thực dương x,y,z thõa mãn \(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}=\sqrt{xyz}\)
Tìm giá trị nhỏ nhất của biểu thức
P=\(\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Cho các số thực dương $x,y,z$ thỏa mãn $x+y+z=1$. Chứng minh rằng:
\(\dfrac{x}{x+\sqrt{x+yz}}+\dfrac{y}{y+\sqrt{y+xz}}+\dfrac{z}{z+\sqrt{z+xy}}\le1\)
\(\sqrt{x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
Cho ba số thực dương x, y, z thỏa mãn: xy+yz+zx=2017. chứng minh : \(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{3}{2}\)
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)