Áp dụng BĐT cosi ta có:

`x^6+y^6+z^6>=3root{3}{x^6y^6z^6}=3x^2y^2z^2`

`=>3x^2y^2z^2<=3`

`=>x^2y^2z^2<=1`

`=>xyz<=1`

`=>(x^3)/(yz)+(y^3)/(zx)+(z^3)/(xy)`

`=(x^4)/(xyz)+(y^4)/(xyz)+(z^4)/(xyz)>=x^4+y^4+z^4(@)`

Áp dụng BĐT bunhia với 2 cặp số `(x^2,y^2,z^2),(x,y,z)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=(x^3+y^3+z^3)^3`

Mà `(x^3+y^3+z^3)^2>=3(x^3y^3+y^3z^3+z^3x^3)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=3(x^3y^3+y^3z^3+z^3x^3)(@@)`

Áp dụng BĐT cosi ta có:

`x^6+1+1>=3root{3}{x^6}=3x^2`

`y^6+1+1>=3y^2`

`z^6+1+1>=3z^2`

`=>x^6+y^6+z^6+6>=3(x^2+y^2+z^2)`

`=>9>=3(x^2+y^2+z^2)`

`=>x^2+y^2+z^2<=3`

Kết hợp với `(@@)`

`=>(x^2+y^2+z^2)(x^4+y^4+z^4)>=(x^2+y^2+z^2)(x^3y^3+y^3z^3+z^3x^3)`

`=>x^4+y^4+z^4>=x^3y^3+y^3z^3+z^3x^3`

Kếp hợp với `(@)`

`=>(x^3)/(yz)+(y^3)/(zx)+(z^3)/(xy)>=x^3y^3+y^3z^3+z^3x^3`

Dấu = xảy ra khi `x=y=z=1`

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.

Nhã Doanh giúp mk vs

sử đề lại đi

sửa đề

\(x^2+y^2+z^2\ge xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(x^2-2zx+z^2\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\left(luôn.đúng\right)\)

Dấu "=' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\)

\(\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

Dấu "=" xảy ra khi \(x=y=z\)

\(VT=\sum\sqrt{\frac{1}{2}\left(x^2+2xy+y^2\right)+\frac{3}{2}\left(x^2+y^2\right)}\)

\(VT\ge\sum\sqrt{\frac{1}{2}\left(x+y\right)^2+\frac{3}{4}\left(x+y\right)^2}=\sum\sqrt{\frac{5}{4}\left(x+y\right)^2}\)

\(VT\ge\frac{\sqrt{5}}{2}\left(x+y\right)+\frac{\sqrt{5}}{2}\left(y+z\right)+\frac{\sqrt{5}}{2}\left(z+x\right)\)

\(VT\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)

\(VT=\left(x^4\right)^2+\left(y^4\right)^2+\left(z^4\right)^2\ge\frac{1}{3}\left(x^4+y^4+z^4\right)^2\)

\(VT\ge\frac{1}{27}\left(x^2+y^2+z^2\right)^4=\frac{1}{27}\left(x^2+y^2+z^2\right)^3\left(x^2+y^2+z^2\right)\)

\(VT\ge\frac{1}{27}\left(3\sqrt[3]{x^2y^2z^2}\right)^3\left(xy+yz+zx\right)=x^2y^2z^2\left(xy+yz+zx\right)\)

Dấu "=" xảy ra khi \(x=y=z\)

Lời giải:

Ta thấy:

\(x^2+xy+y^2=\frac{3}{4}(x^2+2xy+y^2)+\frac{1}{4}(x^2-2xy+y^2)=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\)

\(\geq \frac{3}{4}(x+y)^2\) với mọi $x,y>0$
\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}}{2}(x+y)\)

Hoàn toàn tương tự:

\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}(y+z); \sqrt{z^2+zx+x^2}\geq \frac{\sqrt{3}}{2}(x+z)\)

Cộng theo vế các BĐT trên và rút gọn:

\(\Rightarrow \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\geq \sqrt{3}(x+y+z)\)

Ta có đpcm.

Dấu "=" xảy ra khi $x=y=z$

\(2x^2+2y^2\ge4xy\)

\(4x^2+z^2\ge4xz\)

\(4y^2+z^2\ge4yz\)

Cộng vế:

\(2\left(3x^2+3y^2+z^2\right)\ge4\left(xy+yz+zx\right)\ge20\)

\(\Rightarrow3x^2+3y^2+z^2\ge10\)

Dấu "=" xảy ra tại \(\left(x;y;z\right)=\left(1;1;2\right);\left(-1;-1;-2\right)\)