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\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
Cho x, y, z > 0 và xyz=1. CMR :
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+z}\ge\dfrac{3}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)
áp dụng
\(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2};\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{1}{2}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow A\ge\dfrac{[\left(x+y\right)^2}{2}+z^2].\left(\dfrac{1}{2}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2+\dfrac{1}{z^2}\right)\)
áp dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
\(\Rightarrow A\ge[\dfrac{\left(x+y\right)^2}{2}+z^2].\left(\dfrac{1}{2}.\left(\dfrac{4}{x+y}\right)^2+\dfrac{1}{z^2}\right)=[\dfrac{\left(x+y\right)^2}{2}+z^2].\left(\dfrac{8}{\left(x+y\right)^2}+\dfrac{1}{z^2}\right)=4+1+\dfrac{\left(x+y\right)^2}{2z^2}+\dfrac{8z^2}{\left(x+y\right)^2}=5+\left(\dfrac{\left(x+y\right)^2}{2z^2}+\dfrac{z^2}{2\left(x+y\right)^2}\right)+\dfrac{15z^2}{2\left(x+y\right)^2}\ge5+2.\sqrt{\dfrac{1}{2}.\dfrac{1}{2}}+\dfrac{15\left(x+y\right)^2}{2.\left(x+y\right)^2}=5+1+\dfrac{15}{2}=\dfrac{27}{2}\)
dbxr<=>y=x=z/2>0
\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)
=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)
\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)
và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)
=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)
=>VT>=5+1/2+1/2+15/2=27/2
Áp dụng BĐT cosi:
\(\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge2\sqrt{\dfrac{x^2\left(y+z\right)}{4\left(y+z\right)}}=\dfrac{2x}{2}=x\)
Cmtt \(\dfrac{y^2}{x+z}+\dfrac{x+z}{4}\ge y;\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}+\dfrac{2\left(x+y+z\right)}{4}\ge x+y+z\\ \Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge x+y+z-\dfrac{x+y+z}{2}=\dfrac{x+y+z}{2}\)
Dấu \("="\Leftrightarrow x=y=z\)
\(\sum\dfrac{1}{x}\cdot\sum\dfrac{x}{y^2}\ge\sum^2\dfrac{1}{x}\)(bunhia)