Theo nguyên lý Dirichlet, trong 3 số \(x^2;y^2;z^2\) luôn có ít nhất 2 số cùng phía so với 1

Không mất tính tổng quát, giả sử đó là \(x^2\) và \(y^2\)

\(\Rightarrow\left(x^2-1\right)\left(y^2-1\right)\ge0\)

\(\Leftrightarrow x^2y^2+1\ge x^2+y^2\)

\(\Leftrightarrow x^2y^2+5x^2+5y^2+25\ge6x^2+6y^2+24\)

\(\Leftrightarrow\left(x^2+5\right)\left(y^2+5\right)\ge6\left(x^2+y^2+4\right)\)

\(\Rightarrow\left(x^2+5\right)\left(y^2+5\right)\left(z^2+5\right)\ge6\left(x^2+y^2+4\right)\left(z^2+5\right)\)

\(=6\left(x^2+y^2+1+3\right)\left(1+1+z^2+3\right)\)

\(\ge6\left(x+y+z+3\right)^2\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có:

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

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

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

\(\Rightarrow x^3+y^3+z^3=3xyz\)

\(\Rightarrow5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=15xyz\left(x^2+y^2+z^2\right)\)

Mặt khác:

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5=-z^5\)

\(\Rightarrow x^5+y^5+z^5+5xy\left(x^3+2x^2y+2xy^2+y^3\right)=0\)

\(\Rightarrow x^5+y^5+z^5+\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]=0\)

\(\Rightarrow x^5+y^5+z^5+\left(x+y\right)\left(x^2+xy+y^2\right)=0\)

\(\Rightarrow x^5+y^5+z^5-5xyz\left(x^2+xy+y^2\right)=0\)

\(\Rightarrow2\left(x^5+y^5+z^5\right)-5xyz\left[\left(x^2+2xy+y^2\right)+x^2+y^2\right]=0\)

\(\Rightarrow2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)

Khi đó:\(6\left(x^5+y^5+z^5\right)=15xyz\left(x^2+y^2+z^2\right)=VT\)

\(\Rightarrowđpcm\)

zZz Cool Kid zZz mình chưa hiểu lắm

Bn giải rõ ra dc ko

Lời giải:

Khai triển:

\(\text{VT}=5(x^5+y^5+z^5)+5\underbrace{[x^3(y^2+z^2)+y^3(x^2+z^2)+z^3(x^2+y^2)]}_{M}\)

Xét riêng $M$ kết hợp với điều kiện $x+y+z=0$ ta có

\(M=x^2y^2(x+y)+y^2z^2(y+z)+z^2x^2(x+z)=-(x^2y^2z+y^2z^2x+z^2x^2y)\)

\(\Leftrightarrow M=-xyz(xy+yz+xz)=\frac{-1}{2}xyz[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}xyz(x^2+y^2+z^2)\)

Ta biết đến một hằng thức rất quen thuộc: Nếu $x+y+z=0$ thì \(x^3+y^3+z^3=3xyz\)

Cách chứng minh: \(x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=0-3(-x)(-y)(-z)=3xyz\)

Do đó \(M=\frac{1}{6}(x^3+y^3+z^3)(x^2+y^2+z^2)=\frac{\text{VT}}{30}\)

\(\Rightarrow \text{VT}=5(x^5+y^5+z^5)+5M=5(x^5+y^5+z^5)+\frac{\text{VT}}{6}\)

\(\Rightarrow \text{VT}=6(x^5+y^5+z^5)\) (đpcm)

b) Theo phần a)

\(\left\{\begin{matrix} M=\frac{1}{2}xyz(x^2+y^2+z^2)\\ M=\frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}\end{matrix}\right.\Rightarrow \frac{5(x^2+y^2+z^2)(x^3+y^3+z^3)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)

Mà \(5(x^2+y^2+z^2)(x^3+y^3+z^3)=6(x^5+y^5+z^5)\Rightarrow \frac{6(x^5+y^5+z^5)}{30}=\frac{xyz(x^2+y^2+z^2)}{2}\)

\(\Leftrightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)\) (đpcm)

b)Vì x+y+z=0
=>x+y=-z =>(x+y)^5=-z^5
hay x^5+y^5+5(x^4y+xy^4+2x³y²+2x²y³+)=-z^5
<=>x^5+y^5+z^5+5xy(x³+y³+2x²y+2x²y)=0
<=>x5+y^5+z^5+5xy(x+y)(x²-xy+y²+2xy)=0
<=>x^5+y^5+z^5-5xyz(x²+xy+y²)=0
<=>x^5+y^5+z^5=5xyz(x²+xy+y²)
<=>2(x^5+y^5+z^5)=5xyz(2x²+2xy+2y²)
<=>2(x^5+y^5+z^5)=5xyz[x²+y²+(x+y)²]
<=>2(x^5+y^5+z^5)=5xyz(x³+y²+z²)

Vì bài dài nên mình sẽ tách ra nhé.

1a. Ta có:

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

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

$=-3(-z)(-x)(-y)=3xyz$

$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$

------------------------

$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$

$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$

$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$

$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$

$=-z^5+5xyz^3-5x^2y^2z$

$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$

$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$

Từ $(1);(2)$ ta có đpcm.

1b.

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$

$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$

Do đó:

$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$

$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$

$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$

$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$

$=7xyz(x^2y^2-2xyz^2+z^4)$

$=7xyz(xy-z^2)$

$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$

$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$

$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)

1c. Sử dụng kq phần a,b:

\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)

\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)

\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)

\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)

(đpcm)

1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$

$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)