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Áp dụng bđt phụ \(\dfrac{1}{A+B}\le\dfrac{1}{4}\left(\dfrac{1}{A}+\dfrac{1}{B}\right)\forall A,B>0\)
\(\dfrac{1}{2x+y+z}=\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\) Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Rightarrow\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=1\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{4}\)
Này Nguyễn Trọng Chiến, mk ko hiểu cái chỗ \(\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)??? Sao suy ra được vậy bn??
\(\dfrac{1}{x+x+y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Tương tự: \(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\) ; \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{4}\)
Mk ko hiểu cái dòng đầu Nguyễn Việt Lâm Giáo viên, bn có thể nói chi tiết cách phân tích cho mk đc ko??
Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
Đặt \(Q=x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge x+y+z+\dfrac{9}{x+y+z}\)
\(=x+y+z+\dfrac{1}{x+y+z}+\dfrac{8}{x+y+z}\)
Áp dụng BĐT Cô - si có :
\(\left(x+y+z\right)+\dfrac{1}{x+y+z}\ge2\sqrt{\left(x+y+z\right)\cdot\dfrac{1}{x+y+z}}=2\)
Do \(x+y+z\le1\Rightarrow\dfrac{8}{x+y+z}\ge8\)
Do đó : \(Q\ge8+2=10\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
\(x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge x+y+z+\dfrac{9}{x+y+z}\)
\(VT\ge x+y+z+\dfrac{1}{x+y+z}+\dfrac{8}{x+y+z}\ge2\sqrt{\dfrac{x+y+z}{x+y+z}}+\dfrac{8}{1}=10\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)
\(x,y,z>0\)
Áp dụng BĐT Caushy cho 3 số ta có:
\(x^3+y^3+z^3\ge3\sqrt[3]{x^3y^3z^3}=3xyz\ge3.1=3\)
\(P=\dfrac{x^3-1}{x^2+y+z}+\dfrac{y^3-1}{x+y^2+z}+\dfrac{z^3-1}{x+y+z^2}\)
\(=\dfrac{\left(x^3-1\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)}+\dfrac{\left(y^3-1\right)^2}{\left(x+y^2+z\right)\left(y^3-1\right)}+\dfrac{\left(z^3-1\right)^2}{\left(x+y+z^2\right)\left(x^3-1\right)}\)
Áp dụng BĐT Caushy-Schwarz ta có:
\(P\ge\dfrac{\left(x^3+y^3+z^3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}\)
\(\ge\dfrac{\left(3-3\right)^2}{\left(x^2+y+z\right)\left(x^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)+\left(x+y^2+z\right)\left(y^3-1\right)}=0\)
\(P=0\Leftrightarrow x=y=z=1\)
Vậy \(P_{min}=0\)
Từ \(x\left(\dfrac{1}{y}+\dfrac{1}{z}\right)+y\left(\dfrac{1}{z}+\dfrac{1}{x}\right)+z\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=-2\) ta có:
\(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2+2xyz=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\).
Không mất tính tổng quát, giả sử x + y = 0
\(\Leftrightarrow x=-y\)
\(\Leftrightarrow x^3=-y^3\).
Kết hợp với \(x^3+y^3+z^3=1\) ta có \(z^3=1\Leftrightarrow z=1\).
Vậy \(P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{-y}+\dfrac{1}{y}+\dfrac{1}{1}=1\).
Áp dụng BĐT : \(\dfrac{1}{x}+\dfrac{1}{y}\) \(\geq \) \(\dfrac{4}{x+y}\) \(\Leftrightarrow\) \(\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) \(\geq\) \(\dfrac{1}{x+y}\)
Ta có: \(\dfrac{1}{2x+y+z}\)=\(\dfrac{1}{\left(x+y\right)+\left(x+z\right)}\)\(\leq\)\(\dfrac{1}{4}.\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\)\(\leq\)\(\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{4}\left(\dfrac{1}{x+z}\right)\right)\)=\(\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)(1)
Chứng minh tương tự,ta có:
\(\dfrac{1}{x+2y+z}\) \(\leq\) \(\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\)(2)
\(\dfrac{1}{x+y+2z}\) \(\leq\) \(\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)(3)
Đặt: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\) là VT
Cộng các BĐT(1),(2),(3) lại với nhau ta được:
VT \(\leq\)\(\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)+\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
\(\Leftrightarrow\) VT \(\leq\) \(\dfrac{1}{16}\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)\)=\(\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)=\(\dfrac{1}{4}.4=1\)
\(\Leftrightarrow\) \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\) \(\leq\) 1
Dấu = xảy ra khi x=y=z=\(\dfrac{3}{4}\)
\(P=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\)
\(P=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\)
Áp dụng bđt Cauchy-Schwarz dạng Engel ta có:
\(P\le3-\dfrac{\left(1+1+1\right)^2}{x+1+y+1+z+1}\le3-\dfrac{9}{1+3}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)