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\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)

Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)

Là xong.

Ta có \(xy+xz+yz\le\frac{\left(x+y+z\right)^2}{3}\)

\(\Rightarrow x+y+z+\frac{\left(x+y+z\right)^2}{3}\ge6\)

\(\Rightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)

\(\Rightarrow\left(x+y+z+6\right)\left(x+y+z-3\right)\ge0\)

\(\Rightarrow x+y+z-3\ge0\) (do \(x+y+z+6>0\))

\(\Rightarrow x+y+z\ge3\)

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

Dấu "=" xảy ra khi \(x=y=z=1\)

//Hoặc cách khác sử dụng AM-GM:

\(x^2+1\ge2x\) ; \(y^2+1\ge2y\); \(z^2+1\ge2z\);

\(x^2+y^2+z^2\ge xy+xz+yz\Rightarrow2x^2+2y^2+2z^2\ge2xy+2xz+2yz\)

Cộng vế với vế của 4 BĐT trên ta có:

\(3x^2+3y^2+3z^2+3\ge2\left(x+y+z+xy+xz+yz\right)=12\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)

\(\Rightarrow x^2+y^2+z^2\ge3\)

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.

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

=> \(2x^2+2y^2+2x^2=2xy+2yz+2zx\) 

=> \(2x^2+2y^2+2x^2-2xy-2yz-2zx=0\) 

=> \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\) 

=> x -y =0 ; y - z=0 ; z - x=0

=> x =y; y =z; z=x

=> x=y=z

Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\)

Tương tự: \(y^2+1\ge2y;z^2+1\ge2z\)

\(x^2+y^2\ge2xy\)  \(y^2+z^2\ge2yz\) \(z^2+x^2\ge2zx\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12\)

\(\Leftrightarrow x^2+y^2+z^2\ge3\)

Dấu bằng xảy ra khi x=y=z=1

làm hơi tắt thông cảm

\(A=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\)

\(\Leftrightarrow A^2=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow2A^2=\left(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\right)+\left(\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\right)+\left(\frac{x^2y^2}{z^2}+\frac{z^2x^2}{y^2}\right)+12\)

\(\ge2\left(x^2+y^2+z^2\right)+12=6+12=18\)

\(\Rightarrow A\ge3\)

Đặt vế trái là P

\(P=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\Rightarrow P^2=\frac{x^4y^4+y^4z^4+z^4x^4+2x^2y^2z^2\left(x^2+y^2+z^2\right)}{x^2y^2z^2}\)

\(P^2\ge\frac{x^2y^2z^2\left(x^2+y^2+z^2\right)+6x^2y^2z^2}{x^2y^2z^2}=\frac{9x^2y^2z^2}{x^2y^2z^2}=9\)

\(\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(x=y=1=z\)

x+y+z=3 ms lm đc

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{yz}}{x}+\frac{\left(x+y\right)\sqrt{xy}}{z}+\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\)

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)