Áp dụng bđt AM - GM:

\(x^3+1+1\ge3x;y^3+1+1\ge3y;z^3+1+1\ge3z;2x+2y+2z\ge6\sqrt[3]{xyz}=6\).

Cộng vế với vế các bđt trên rồi rút gọn ta có đpcm.

Áp dụng BĐT Cosi:

\(\left(x^3+1+1\right)+\left(y^3+1+1\right)+\left(z^3+1+1\right)\)

\(\ge3\left(x+y+z\right)\)

\(\ge x+y+z+2.3\sqrt[3]{xyz}\)

\(=x+y+z+6\)

\(\Rightarrow x^3+y^3+z^3\ge x+y+z\)

Đẳng thức xảy ra khi \(x=y=z=1\)

Bài 3:

a, (\(x\)+y+z)²

=((\(x\)+y) +z)²

= (\(x\) + y)² + 2(\(x\) + y)z + z²

= \(x^2\) + 2\(xy\) + y² + 2\(xz\) + 2yz + z²

=\(x^2\) + y² + z² + 2\(xy\) + 2\(xz\) + 2yz

b, (\(x-y\))(\(x^2\) + y² + z² - \(xy\) - yz - \(xz\))

= \(x^3\) + \(xy^2\) + \(xz^2\) - \(x^2\)y - \(xyz\) - \(x^2\)z - y³ 

Đến dây ta thấy xuất hiện \(x^3\) - y³ khác với đề bài, em xem lại đề bài nhé

c,

(\(x\) + y + z)³ 

=(\(x\) + y)³ + 3(\(x\) + y)²z + 3(\(x\)+y)z² + z³

= \(x^3\) + 3\(x^2\)y + 3\(xy^{2^{ }}\) + y³ +  3(\(x\)+y)z(\(x\) + y + z) + z³

= \(x^3\) + y³ + z³ + 3\(xy\)(\(x\) + y) + 3(\(x+y\))z(\(x+y+z\))

= \(x^3\) + y³ + z³ + 3(\(x\) + y)( \(xy\) + z\(x\) + yz + z²)

= \(x^3\) + y³ + z³ + 3(\(x\) + y){(\(xy+xz\)) + (yz + z²)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y){ \(x\)( y +z) + z(y+z)}

= \(x^3\) + y³ + z³ + 3(\(x\) + y)(y+z)(\(x+z\)) (đpcm)

Ta có: \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-\left[3xy\left(x+y\right)+3xyz\right]\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-\left[3xy\left(x+y+z\right)\right]\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-zy+z^2-3xy\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)(đpcm)

Ta có: \(x^3-y^3=3x-3y\Leftrightarrow x^2+xy+y^2=3\) (Do \(x\neq y\)).

Tương tự: \(y^2+yz+z^2=3;z^2+zx+x^2=3\).

Cộng vế với vế ta có: \(2\left(x^2+y^2+z^2\right)+xy+yz+zx=9\)

\(\Leftrightarrow\dfrac{3\left(x^2+y^2+z^2\right)}{2}+\dfrac{\left(x+y+z\right)^2}{2}=9\).

Mặt khác, từ đó ta cũng có: \(\left(x^2+xy+y^2\right)-\left(y^2+yz+z^2\right)=0\Leftrightarrow\left(x+y+z\right)\left(x-z\right)=0\Leftrightarrow x+y+z=0\).

Do đó \(x^2+y^2+z^2=6\left(đpcm\right)\).

Cặc

Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
 

( x + y + z)³ - x³ - y³ - z³=x³+y³+z³+3(a+b)(a+c)(b+c)- x³ - y³ - z³

                                              = 3(a+b)(b+c)(a+c)

\(\left(x+y-z\right)^3-x^3-y^3+z^3\)

\(=\left[\left(x+y\right)-z\right]^3-x^3-y^3+z^3\)

\(=\left(x+y\right)^3-z^3-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)

\(=x^3+y^3-z^3+3xy\left(x+y\right)-3\left(x+y\right)z\left(x+y-z\right)-x^3-y^3+z^3\)

\(=3xy\left(x+y\right)-3z\left(x+y\right)\left(x+y-z\right)\)

\(=3\left(x+y\right)\left[xy-z\left(x+y-z\right)\right]\)

\(=3\left(x+y\right)\left(xy-zx-yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]\)

\(=3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)

#\(Urushi\text{☕}\)

Áp dụng (a+b)3 = a3+b3+3ab(a+b), ta có:

(x+y+z)3-x3-y3-z3

=[(x+y)+z]3-x3-y3-z3

=(x+y)3+z3+3z(x+y)(x+y+z)-x3-y3-z3

=x3+y3+3xy(x+y)+z3+3z(x+y)(x+y+z)-x3-y3-z3

=3(x+y)(xy+xz+yz+z2)

=3(x+y)[x(y+z)+z(y+z)]

=3(x+y)(y+z)(x+z)

=(x+y-z-x)[(x+y-z)^2+x(x+y-z)+x^2]-(y-z)(y^2+yz+z^2)

=(y-z)(x^2+y^2+z^2+2xy-2xz-2yz+x^2+xy-xz+x^2-y^2-yz-z^2)

=(y-z)(3x^2+3xy-3xz-3yz)

=3(y-z)(x^2+xy-xz-yz)

=3(y-z)[x(x+y)-z(x+y)]

=3(y-z)(x+y)(x-z)

Ta rút gọn tử thức trc: \(x^3+y^3+z^3-3xyz=x^3+y^3+z^3+x^2y-x^2y+xy^2-xy^2+y^2z-y^2z+yz^2-yz^2+x^2z-x^2z+xz^2-xz^2-xyz-xyz-xyz=x^2\left(x+y+z\right)+y^2\left(x+y+z\right)+z^2\left(x+y+z\right)-x\left(x+y+z\right)-yz\left(x+y+z\right)-xz\left(x+y+z\right)=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=\frac{1}{2}\left(x+y+z\right)\left(x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2\right)=\frac{1}{2}\left(x+y+z\right)\left(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right)\)tới đây rút gọn đc rồi chứ