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

\(b,x^3+y^3+z^3-3xyz\\ =\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xz-yz+2xy-3xy\right)\\ =0\left(x^2+y^2+z^2-xz-yz-xy\right)=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

\(\left(3^x;3^y;3^z\right)=\left(a;b;c\right)\Rightarrow\left\{{}\begin{matrix}a;b;c>0\\ab+bc+ca=abc\end{matrix}\right.\)

BĐT cần chứng minh trở thành:

\(\dfrac{a^2}{a+bc}+\dfrac{b^2}{b+ca}+\dfrac{c^2}{c+ab}\ge\dfrac{a+b+c}{4}\)

Thật vậy, ta có:

\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

\(VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+c\right)\left(b+c\right)}\)

Áp dụng AM-GM:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge\dfrac{3a}{4}\)

Làm tương tự với 2 số hạng còn lại, cộng vế với vế rồi rút gọn, ta sẽ có đpcm

Một bài toán "lừa" người ta:

Đặt \(a=x-y,b=y-z,c=z-x\Rightarrow a+b+c=0\).

Ta có hằng đẳng thức \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

Trong trường hợp này thì \(a+b+c=0\) nên suy ra đpcm.

Hình như đề sai rồi

đúng đề mà bạn

Vậy b nói xem thử khi nào nó = 4.

a, x^4 - 5x^2 + 4

= x^4 - 4x^2- x+ 4

= x^2  . (x^2 - 4) - (x^2 - 4)

= (x^2 - 4) . (x^2 - 1)

= (x - 2) . (x + 2) . (x - 1) . (x + 1)

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

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

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

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

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

\(=3x^2y+3xy^2+3x^2z+6xyz+3y^2z+3xz^2+3yz^2\) (1)

• \(VP=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

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

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

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

\(=6xyz+3x^2y+3xz^2+3x^2z+3y^2z+3xy^2+3yz^2\) (2)

Từ (1) và (2) suy ra \(VT=VP\) (đpcm)

Ta có: \(\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)

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

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

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

\(=x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\left(\forall x,y,z\right)\)

=> đpcm

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

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

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

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

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

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

Chúc bạn học tốt 

T I C K nha cảm ơn bạn