cho x,y,z >0 và x+y+z=1
tìm Min \(P=\sqrt{x^2+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{x^2}}\)
cho x,y,z>0 x+y+z<=1
tìm GTNN:P= \(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
Sử dụng bất đẳng thức Minkovski, ta có:
\(P = \sqrt {{{\left( {x + y + z} \right)}^2} + {{\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)}^2}} \)
\( \ge \sqrt {\left[ {{{\left( {x + y + z} \right)}^2} + \frac{1}{{{{\left( {x + y + z} \right)}^2}}}} \right] + \frac{{80}}{{{{\left( {x + y + z} \right)}^2}}}} \)
\(\ge \sqrt{2+\dfrac{80}{1}} =\sqrt{82}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1}{3}.\)
Kết luận ...
\(\sqrt{x^2+\dfrac{1}{x^2}}=\dfrac{1}{\sqrt{82}}\sqrt{\left(1^2+9^2\right)\left(x^2+\dfrac{1}{x^2}\right)}\ge\dfrac{1}{\sqrt{82}}\left(x+\dfrac{9}{x}\right)\)
tương tự với \(\sqrt{y^2+\dfrac{1}{y^2}};\sqrt{z^2+\dfrac{1}{z^2}}\)
\(=>P\ge\dfrac{1}{\sqrt{81}}\left(x+\dfrac{9}{x}+y+\dfrac{9}{y}+z+\dfrac{9}{z}\right)\)
có \(x+\dfrac{9}{x}=x+\dfrac{1}{9x}+\dfrac{80}{9x}\ge2\sqrt{\dfrac{1}{9}}+\dfrac{80}{9x}\)
tương tự với \(y+\dfrac{9}{y};z+\dfrac{9}{z}\)
\(=>P\ge\dfrac{1}{\sqrt{82}}\left[2\sqrt{\dfrac{1}{9}}.3+\dfrac{\left(\sqrt{80}+\sqrt{80}+\sqrt{80}\right)^2}{9\left(x+y+z\right)}\right]=\dfrac{1}{\sqrt{82}}.82=\sqrt{82}\)
dấu"=" xảy ra<=>x=y=z=1/3
cho x,y,z>0 và x+y+z=\(\dfrac{3}{2}\)
tìm Min \(P=\dfrac{\sqrt{x^2+xy+y^2}}{\left(x+y\right)^2+1}+\dfrac{\sqrt{y^2+yz+z^2}}{\left(y+z\right)^2+1}+\dfrac{\sqrt{z^2+zx+x^2}}{\left(z+x\right)^2+1}\)
Đề bài sai, biểu thức này ko có min
Cho 0≤x,y,z≤1
Tìm max \(D=\sqrt{\dfrac{x}{1+yz}}+\sqrt{\dfrac{y}{1+zx}}+\dfrac{z}{2+2xy}\)
\(D\le\dfrac{1}{2}\left(1+\dfrac{x}{1+yz}\right)+\dfrac{1}{2}\left(1+\dfrac{y}{1+zx}\right)+\dfrac{z}{2+2xy}\)
\(=1+\dfrac{x}{2\left(1+yz\right)}+\dfrac{y}{2\left(1+zx\right)}+\dfrac{z}{2\left(1+xy\right)}\)
Do \(0\le x;y;z\le1\)
\(\Rightarrow\left(1-x\right)\left(1-y\right)\ge0\Leftrightarrow xy+1\ge x+y\)
\(\Leftrightarrow2\left(xy+1\right)\ge xy+1+x+y\ge x+y+z\)
\(\Rightarrow\dfrac{z}{2\left(1+xy\right)}\le\dfrac{z}{x+y+z}\)
Tương tự: \(\dfrac{x}{2\left(1+yz\right)}\le\dfrac{x}{x+y+z}\) ; \(\dfrac{y}{2\left(1+zx\right)}\le\dfrac{y}{x+y+z}\)
Cộng vế:
\(P\le1+\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}=2\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\)
\(a+b+c+2=abc\)
\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1=ab+bc+ca+2\left(a+b+c\right)+3\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(a+1\right)\left(b+1\right)+\left(b+1\right)\left(c+1\right)+\left(c+1\right)\left(a+1\right)\)
\(\Leftrightarrow\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}=1\)
Ta có:
\(a^2+a^2+4\ge\dfrac{1}{3}\left(a+a+2\right)^2=\dfrac{4}{3}\left(a+1\right)^2\)
\(\Rightarrow\sum\dfrac{a+1}{a^2+2}\le\dfrac{3}{2}\sum\dfrac{a+1}{\left(a+1\right)^2}=\dfrac{3}{2}\sum\dfrac{1}{a+1}=\dfrac{3}{2}\)
Cho x,y,z>0 thỏa mãn xyz=1. Tìm min \(P=\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\dfrac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\dfrac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
Áp dụng bất đẳng thức cauchy:
\(P=\sum\dfrac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}\ge\sum\dfrac{2x^2\sqrt{yz}}{y\sqrt{y}+2z\sqrt{z}}=\sum\dfrac{2\sqrt{x^3}\sqrt{xyz}}{\sqrt{y^3}+2\sqrt{z^3}}=\sum\dfrac{2\sqrt{x^3}}{\sqrt{y^3}+2\sqrt{z^3}}\)(vì xyz=1).
đặt \(\left\{{}\begin{matrix}\sqrt{x^3}=a\\\sqrt{y^3}=b\\\sqrt{z^3}=c\end{matrix}\right.\)(\(a,b,c>0\))thì giả thiết trở thành cho abc=1. tìm Min \(P=\dfrac{2a}{b+2c}+\dfrac{2b}{c+2a}+\dfrac{2c}{a+2b}\)
Áp dụng BĐT cauchy-schwarz:
\(P=2\left(\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\right)\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\)( AM-GM \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\))
Dấu = xảy ra khi a=b=c=1 hay x=y=z=1
Cho x,y,z>0 và x+y+z=1.CMR:\(\sqrt{x^2+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{z^2}}\sqrt{z^2+\dfrac{1}{x^2}}>=\sqrt{82}\)
Cho x,y,z > 0 và \(x+y+z\le\dfrac{3}{2}\). CMR :
\(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{3}{2}\sqrt{17}\)
Cho x, y, z > 0 thoả mãn x+y+z=1. Chứng minh rằng:
a) \(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\sqrt{82}\)
b) \(\sqrt{x^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}\ge\sqrt{163}\)
c)\(\sqrt{x^2+\dfrac{2}{y^2}+\dfrac{3}{z^2}}+\sqrt{y^2+\dfrac{2}{z^2}+\dfrac{3}{x^2}}+\sqrt{z^2+\dfrac{2}{z^2}+\dfrac{3}{y^2}}\ge\sqrt{406}\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Cho x,y,z>0 và x+y+z = xyz
CMR
\(\dfrac{1}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\le\dfrac{3}{2}\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow ab+bc+ca=1\)
Đặt vế trái là P, ta có:
\(P=\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\)
\(P=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}+\dfrac{b}{\sqrt{b^2+ab+bc+ca}}+\dfrac{c}{\sqrt{c^2+ab+bc+ca}}\)
\(P=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)+\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)+\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)