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\(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Leftrightarrow\begin{matrix}a+b+c=1\\a^4+b^4+c^4\ge abc\end{matrix}\) \(x,y,z\ne0\Rightarrow a,b,c\ne0\)
\(a^2+b^2+x^2\ge ab+bc+ac\) (*){cơ bản} \(\Rightarrow\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\ge\left(ab.ac\right)+\left(ab.bc\right)+\left(ac.bc\right)=abc\left(a+b+c\right)=abc\)
(*) bình phương hai vế
\(\Leftrightarrow a^4+b^4+c^4+2\left(ab\right)^2+2\left(ac\right)^2+2\left(bc\right)^2\ge\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge-\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]+2abc\ge-abc+2abc=abc=>dpcm\)Đẳng thức:
a=b=c=1/3=> x=y=z=3
ta co \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\) \(\Rightarrow\) \(\dfrac{1}{x.x}+\dfrac{1}{y.y}+\dfrac{1}{z.z}=1\)
\(\Rightarrow\dfrac{1}{x.x.x}+\dfrac{1}{y.y.y}+\dfrac{1}{z.z.z}=1\)\(\Rightarrow\dfrac{1}{x.x.x.x}+\dfrac{1}{y.y.y.y}+\dfrac{1}{z.z.z.z}=1\Leftrightarrow\dfrac{1}{x^4}+\dfrac{1^{ }}{y^4}+\dfrac{1}{z^4}=1\)
\(\Rightarrow\)\(\dfrac{1}{x^4}+\dfrac{1}{y^4}+\dfrac{1}{z^4}\)>= \(\dfrac{1}{x.y.z}\)
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=\(\dfrac{3}{4}\)
ĐK: \(x,y,z,x+y+z\ne0\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\left(\dfrac{1}{z}-\dfrac{1}{x+y+z}\right)=0\)
\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{xy+yz+zx+z^2}{xyz\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
\(\circledast x=-y\)
\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{1}{-y^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{1}{z^3}\)
\(\dfrac{1}{x^3+y^3+z^3}=\dfrac{1}{-y^3+y^3+z^3}=\dfrac{1}{z^3}\)
Vậy \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{1}{x^3+y^3+z^3}\)
Lầm tương tự với hai trường hợp còn lại ta có đpcm
a: Thiếu vế phải rồi bạn
b: \(\Leftrightarrow\dfrac{x+y}{xy}>=\dfrac{4}{x+y}\)
\(\Leftrightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow\left(x-y\right)^2>=0\)(luôn đúng)
Sửa đề: \(\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\ge\dfrac{3}{4}\)
Đặt \(P=\dfrac{x}{x+1}+\dfrac{y}{y+1}+\dfrac{z}{z+1}\)
\(P=\dfrac{x+1}{x+1}-\dfrac{1}{x+1}+\dfrac{y+1}{y+1}-\dfrac{1}{y+1}+\dfrac{z+1}{z+1}-\dfrac{1}{z+1}\)
\(P=1-\dfrac{1}{x+1}+1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1}\)
\(P=3-\left(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\right)\)
Ta có:
\(\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{9}{x+y+z+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{9}{4}\) ( vì \(x+y+z=1\) )
\(\Rightarrow P\ge3-\dfrac{9}{4}=\dfrac{3}{4}\left(đpcm\right)\)
Dấu "=" xảy ra khi \(x+1=y+1=z+1\)
\(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Vậy \(Max_P=\dfrac{3}{4}\) khi \(x=y=z=\dfrac{1}{3}\)
thanks bạn