Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=xy\left(x+y\right)\)

\(\Rightarrow VT\le\dfrac{1}{xy\left(x+y\right)+xyz}+\dfrac{1}{yz\left(y+z\right)+xyz}+\dfrac{1}{zx\left(z+x\right)+xyz}\)

\(\Rightarrow VT\le\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=\dfrac{1}{x+y+z}.\left(\dfrac{x+y+z}{xyz}\right)=\dfrac{1}{xyz}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z\)

Với x ; y > 0 , cần c/m : \(x^3+y^3\ge xy\left(x+y\right)\)

Ta có : \(x^3+y^3-xy\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-xy\right)=\left(x+y\right)\left(x-y\right)^2\ge0\)

( điều này luôn đúng với mọi x ; y > 0 )

=> BĐT được c/m

Áp dụng vào bài toán , ta có :

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{xz\left(x+z\right)+xyz}=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{xz\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=\frac{1}{xyz}\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z;x,y,z>0\)

Theo bài ra,ta có:

\(xyz=-20\cdot\frac{1}{2}\cdot\frac{1}{5}=-\frac{20}{10}=-2\)

\(\Rightarrow A=-2+\left(-2\right)^2+\left(-2\right)^3+.....+\left(-2\right)^{2019}\)

\(\Rightarrow-2A=\left(-2\right)^2+\left(-2\right)^3+\left(-2\right)^4+....+\left(-2\right)^{2020}\)

\(\Rightarrow-3A=-2^{2020}+2\)

\(\Rightarrow A=\frac{-2^{2020}+2}{-3}\)

x/2=y/3=z/5 và xyz +30

Ta có:

\(x^2+y^2\ge2xy\Rightarrow x^2+y^2-xy\ge xy\)

\(\Leftrightarrow\left(x+y\right)\left(x^2+y^2-xy\right)\ge xy\left(x+y\right)\)

\(\Leftrightarrow x^3+y^3\ge xy\left(x+y\right)\)

\(\Rightarrow\frac{1}{x^3+y^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}=\frac{1}{x+y+z}.\frac{1}{xy}\)

Tương tự: \(\frac{1}{y^3+z^3+xyz}\le\frac{1}{x+y+z}.\frac{1}{yz}\) ;\(\frac{1}{z^3+x^3+xyz}\le\frac{1}{x+y+z}.\frac{1}{zx}\)

\(\Rightarrow\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{z^3+x^3+xyz}\)

\(\le\frac{1}{x+y+z}.\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{x+y+z}{\left(x+y+z\right)xyz}=\frac{1}{xyz}\)

Dấu \(=\) xảy ra \(\Leftrightarrow x=y=z>0\)

AD BĐT X^3+Y^3>=XY(X+Y) LÀ RA

Có BĐT phụ:

\(a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow a^3-a^2b+b^3-ab^2\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)

Áp dụng

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\)

\(\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{zx\left(z+x\right)+xyz}\)

\(=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}\)

\(=\frac{1}{xyz}\)

do x,y,z là các số dương nên

\(x^2-xy+y^2\ge xy\Leftrightarrow x^3+y^3\ge xy\left(x+y\right)\)

tương tự ta cũng có : \(y^3+z^3\ge yz\left(y+z\right)\)

\(z^3+x^3\ge zx\left(z+x\right)\)

\(\Rightarrow\Sigma\dfrac{1}{x^3+y^3+xyz}\le\Sigma\dfrac{1}{xy\left(x+y+z\right)}=\dfrac{1}{x+y+z}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\)

\(=\dfrac{1}{x+y+z}\left(\dfrac{x+y+z}{xyz}\right)=\dfrac{1}{xyz}\left(đpcm\right)\)

4/xyz vì xyz cùng mẫu

k nha

\(\frac{1}{xyz}+\frac{1}{xyz}+\frac{2}{xyz}=\frac{1+1+2}{xyz}=\frac{4}{xyz}\)

\(=xyz^2\)

đúng không

\(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2+\left(-\frac{1}{4}\right)xyz^2\)

=\(\left(\frac{3}{4}+\frac{1}{2}-\frac{1}{4}\right)xyz^2\)

=\(xyz^2\)

\(\frac{3}{4}xyz^2+\frac{1}{2}xyz^2+\left(-\frac{1}{4}\right)xyz^2\)

=\(xyz^2\left[\frac{3}{4}+\frac{1}{2}+\left(-\frac{1}{4}\right)\right]\)

=\(xyz^2.1\)

= \(xyz^2\)