cho x,y,z dương thỏa \(xyz=1\)
tìm min \(P=\dfrac{x+2}{x^3\left(y+z\right)}+\dfrac{y+2}{y^3\left(z+x\right)}+\dfrac{z+2}{z^3\left(x+y\right)}\)
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 là các số thực dương thỏa mãn x+y+z=xyz.CMR
\(\dfrac{x}{1+x^2}+\dfrac{2y}{1+y^2}+\dfrac{3z}{1+z^2}=\dfrac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Cho x, y, z dương thỏa mãn xyz=1. Tìm GTLN của \(\dfrac{1}{\sqrt{\left(x+y\right)^2+\left(x+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(y+z\right)^2+\left(y+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(z+x\right)^2+\left(z+1\right)^2+4}}\)
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
Cho x, y, z đôi một khác nhau thỏa mãn \(\left(x+z\right)\left(y+z\right)=1\). Tìm Min
\(M=\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{\left(x+z\right)^2}+\dfrac{1}{\left(y+z\right)^2}\)
Cho x, y, z dương thỏa mãn xyz = 1. Tìm GTLN:
P = \(\dfrac{1}{\left(3x+1\right)\left(y+z\right)+x}+\dfrac{1}{\left(3y+1\right)\left(z+x\right)+y}+\dfrac{1}{\left(3z+1\right)\left(x+y\right)+z}\)
\(P=\dfrac{1}{3x\left(y+z\right)+x+y+z}+\dfrac{1}{3y\left(z+x\right)+x+y+z}+\dfrac{1}{3z\left(x+y\right)+x+y+z}\)
\(P\le\dfrac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3y\left(z+x\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3z\left(x+y\right)+3\sqrt[3]{xyz}}\)
\(P\le\dfrac{1}{3x\left(y+z\right)+3}+\dfrac{1}{3y\left(z+x\right)+3}+\dfrac{1}{3z\left(x+y\right)+3}\)
Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3\left(b^3+c^3\right)+1}+\dfrac{1}{b^3\left(c^3+a^3\right)+1}+\dfrac{1}{c^3\left(a^3+b^3\right)+1}\right)\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3bc\left(b+c\right)+1}+\dfrac{1}{b^3ac\left(a+c\right)+1}+\dfrac{1}{c^3ab\left(a+b\right)+1}\right)\)
\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{bc}{a\left(b+c\right)+bc}+\dfrac{ac}{b\left(a+c\right)+ac}+\dfrac{ab}{c\left(a+b\right)+ab}\right)=\dfrac{1}{3}\)
\(P_{max}=\dfrac{1}{3}\) khi \(a=b=c=1\) hay \(x=y=z=1\)
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 là số thực dương thỏa mãn xy+yz+xz=xyz
cmr \(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}+\dfrac{xz}{y^3\left(1+x\right)\left(1+z\right)}\ge\dfrac{1}{16}\)
Gọi cái thiệt gớm đó là P
Ta có:
\(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64y}+\dfrac{1+y}{64x}\ge3\sqrt[3]{\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}.\dfrac{1+x}{64y}.\dfrac{1+y}{64x}}=\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{64x}-\dfrac{1}{64y}-\dfrac{1}{32}\left(1\right)\)
Tương tự ta cũng có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{64y}-\dfrac{1}{64z}-\dfrac{1}{32}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{64z}-\dfrac{1}{64x}-\dfrac{1}{32}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta được
\(P\ge\dfrac{3}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
Cho các số x, y, z dương thỏa mãn: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\)
Cmr: \(\dfrac{1}{\left(2x+y+z\right)^2}+\dfrac{1}{\left(2y+z+x\right)^2}+\dfrac{1}{\left(2z+x+y\right)^2}\ge\dfrac{3}{16}\)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
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