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 \(x=y=z\)

x^2 = yz => x/y = z/x 

Theo dãy tỉ số (=)

  \(\frac{x}{y}=\frac{z}{x}=\frac{z+x}{x+y}=\frac{z-x}{x-y}\)

=> \(\frac{z+x}{x+y}=\frac{z-x}{x-y}\Rightarrow\frac{x+y}{x-y}=\frac{z+x}{z-x}\)

Lời giải:

Đặt \((x,y,z)=(a^2,b^2,c^2)\). Bài toán tương đương với:

\(\frac{bc(b+c)}{a}+\frac{ac(a+c)}{b}+\frac{ab(a+b)}{c}\geq 2(a^2+b^2+c^2)\)

Biến đổi ta thấy:

\(\text{VT}=a^2\left ( \frac{b}{c}+\frac{c}{b} \right )+b^2\left ( \frac{a}{c}+\frac{c}{a} \right )+c^2\left ( \frac{a}{b}+\frac{b}{a} \right )\)

Áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} \frac{a}{b}+\frac{b}{a}\geq 2\\ \frac{a}{c}+\frac{c}{a}\geq 2\\ \frac{b}{c}+\frac{c}{b}\geq 2\end{matrix}\right.\Rightarrow \text{VT}\geq 2(a^2+b^2+c^2)=\text{VP}\)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\Leftrightarrow x=y=z>0\)

Áp dụng BĐT AM-GM ta có:

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

Tương tự cho 2 BĐT còn lại thì được:

\(\dfrac{2xy}{z}+\dfrac{2yz}{x}+\dfrac{2xz}{y}\ge2\left(x+y+z\right)\)

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

Tiếp tục dùng AM-GM:

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

Tương tự rồi cộng theo vế có:

\(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\ge x+y+z\) (đúng)

Hay ta có ĐPCM. Khi \(x=y=z\)

Đề này à: \(\dfrac{\left(y+z\right)\sqrt{yz}}{x}+\dfrac{\left(z+x\right)\sqrt{zx}}{y}+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge2\left(x+y+z\right)\)

Dùng máy tính kiểm tra. (đề sai không?)

Thế x=1, y=2, z=3

VT = 17,12576389

VP = 12

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

=> \(2x^2+2y^2+2x^2=2xy+2yz+2zx\) 

=> \(2x^2+2y^2+2x^2-2xy-2yz-2zx=0\) 

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

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

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

=> x=y=z

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.

Ta co : 

\(x^2=\frac{x}{y};yz=\frac{z}{x}\Rightarrow x^2=yz=\frac{x}{y}=\frac{z}{x}\)

Dat : \(\frac{x}{y}=\frac{z}{x}=k\)

x=yk

z=xk

\(\frac{x+y}{x-y}=\frac{yk+y}{yk-y}=\frac{y.\left(k+1\right)}{y.\left(k-1\right)}=\frac{k+1}{k-1}\)        (1)

\(\frac{z+x}{z-x}=\frac{xk+x}{xk-x}=\frac{x.\left(k+1\right)}{x.\left(k-1\right)}=\frac{k+1}{k-1}\)       (2)

​               Vậy  từ (1) và (2) suy ra \(\frac{x+y}{x-y}=\frac{z+x}{z-x}\)