Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) 

\(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) 

Tương tự rồi cộng từng vế, ta có: 

\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{4}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\) 

=> ĐPCM

chtt

Nhanh to cho card 20

6) Ta có

\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)

\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+2xz+yz+2xy+zx+2yz}\)

\(\Leftrightarrow A\ge\frac{1}{3\left(xy+yz+zx\right)}\ge\frac{1}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)

Với 2 số dương bất kì: ( 1 )

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)Vì x và y dương nên \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)

Áp dụng ( 1 ): \(\frac{4}{2x+y+z}=\frac{4}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{x+y}+\frac{1}{x+z}\)

Mà: \(\frac{1}{x+y}+\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{4}\)\(=\frac{1}{4}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Nên: \(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Tương tự ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)

Và \(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

Cộng vế với vế các bất đẳng thức kết hợp với điều kiện \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) nên ta có đpcm

a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)

Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2

b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)

Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)

Giải:

Ta có: x, y, z >0

Áp dụng BĐT Cô si ta có:

\(\left(x+y\right)\ge2\sqrt{xy}\) và \(\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{\frac{1}{xy}}\)

=> \(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\frac{1}{xy}}=4\)

<=> \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\frac{1}{x+y}\le4\left(\frac{1}{x}+\frac{1}{y}\right)\)               (*)

Áp dụng (*) ta có: 

\(\frac{1}{2x+y+z}=\frac{1}{x+y+x+z}=\frac{1}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)\)        (1)

\(\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}=\frac{1}{\left(x+y\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)        (2)

\(\frac{1}{x+y+2z}=\frac{1}{x+z+y+z}=\frac{1}{\left(x+z\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\)        (3)

Cộng 2 vế của (1), (2), (3) ta có

\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\) (đpcm)
 

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