Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

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

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

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

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

giờ nhân cả tử và mẫu mỗi phân thức vs mỗi tử của nó rồi sử dụng BDT bunhiacopxki là ra thôi bn

\(\frac{x^2}{x^3-xyz+2013x}+\frac{y^2}{y^3-xyz+2013y}+\frac{z^2}{z^3-xyz+2013z}\)

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

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

\(VT=\text{Σ}_{cyc}\frac{x}{x^2-yz+2013}=\text{Σ}_{cyc}\frac{x^2}{x^3-xyz+2013x}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)(bđt Cauchy - Schwarz dạng Engel)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)+2013\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx+2013\right)}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3\left(xy+yz+zx\right)+2013\right]}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[\left(x+y+z\right)^2-3.671+2013\right]}\)

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

(Dấu "=" xảy ra khi x = y = z = \(\frac{\sqrt{2013}}{3}\))

Phân tích nhân tử là được

\(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=-y\\y=-z\\z=-x\end{cases}}\)

Với \(x=-y\) thì

\(\hept{\begin{cases}x^{2013}+y^{2013}+z^{2013}=z^{2013}\\\left(x+y+z\right)^{2013}=z^{2013}\end{cases}}\)

\(\Rightarrow x^{2013}+y^{2013}+z^{2013}=\left(x+y+z\right)^{2013}\)

Tương tự cho các trường hợp còn lại.

bt làm rồi hỏi vui thôi ^^

Từ giả thiết , ta có :

( x + y + z)( xy + yz + xz ) = xyz

x( xy + yz + xz) + y( xy + yz + xz ) + z( xy + yz + xz ) - xyz = 0

x²y + xyz + x²z + xy² + y²z + xyz + xyz + yz² + xz² - xyz = 0

x²y + x²z + xy² + y²z + yz² + xz² + 2xyz = 0

xy( x + y) + xz( x + z) + yz( y + z) + 2xyz = 0

xy( x + y + z) + xz( x + y + z) + yz( y + z) = 0

( x + y + z)x( y + z) + yz( y + z) = 0

( y + z)( x² + xy + xz + yz ) = 0

( y + z)[ x( x + y ) + z( x + y) ] = 0

( y + z)( y + x )( x + z) = 0

Suy ra :

* x + y = 0 --> x = - y . Thay vào đẳng thức cần chứng minh , ta có

( - y)²⁰¹³ + y²⁰¹³ + z²⁰¹³ = ( - y + y + z)²⁰¹³

Khi đó , ta có : z²⁰¹³ = z²⁰¹³ , luôn đúng

* Tương tự , thử với các trường hợp khác : y = - z ; x = - z

Vậy , đảng thức được chứng mình

Ta có (x+y+z)(xy+yz+xz)=xyz

<=>\((x+y+z)(\frac{xyz}{z}+\frac{xyz}{y}+\frac{xyz}{x})=xyz \)

<=>(x+y+z)(\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1 \)

<=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z} \)

<=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0 \)

<=>\(\frac{x+y}{xy}+\frac{x+y}{z(x+y+z)} \)

<=>\((x+y)(\frac{1}{xy}+\frac{1}{z(x+y+z)}) \)

<=>\((x+y)(\frac{xz+yz+z^2+xy}{xyz(x+y+z)} \)

<=>\((x+y)(y+z)(x+z)(\frac{1}{xyz(x+y+z)} )\)

=>x=-y

hoặc y=-z

hoặc x=-z

Thay vào Pt => đpcm

\(\frac{x}{x^2-yz+2013}+\frac{y}{y^2-zx+2013}+\frac{z}{z^2-xy+2013}\)

\(=\frac{1}{\frac{x^2-yz+2013}{x}}+\frac{1}{\frac{y^2-zx+2013}{y}}+\frac{1}{\frac{z^2-xy+2013}{z}}\)

\(=\frac{1}{x+3y+3z+\frac{2yz}{x}}+\frac{1}{y+3z+3x+\frac{2xz}{y}}+\frac{1}{z+3x+3y+\frac{2xy}{z}}\)

\(\ge\frac{9}{7\left(x+y+z\right)+2xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{7\left(x+y+z\right)+2xyz\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}=\)

\(=\frac{9}{7\left(x+y+z\right)+2xyz.\frac{1}{xyz}.\left(x+y+z\right)}=\frac{9}{9\left(x+y+z\right)}=\frac{1}{x+y+z}\)

Ta có đpcm

bó tay rùi bạn !!!! ~_~

65756578687696453724756545345363637635754754695622534434

\(VT=\frac{x^2}{x^3-xyz-2013x}+\frac{y^2}{y^3-xyz-2013y}+\frac{z^2}{z^3-xyz-2013z}\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\right]}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)=VP

đúng rồi ạ nhưng chỉ cần c/m đẳng thức phụ như thế này thôi ạ\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\) =>\(\frac{\left(a+b\right)2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) hay \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) là xong

2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)

Còn câu 1 nữa ạ, ai giải giúp em vớii