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\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\\ =\frac{x}{y-z}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right)\\ =\frac{x}{\left(y-x\right)^2}=-\left(\frac{y}{z-x}+\frac{z}{x-y}\right).\frac{1}{y-x}=\frac{-xy+y^2-z^2+xz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(1\right)\)
Tự làm với 2 phân thức còn lại, ta có:
\(\frac{y}{\left(z-x\right)^2}=\frac{-x^2+z^2+xy-yz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(2\right)\)
\(\frac{z}{\left(x-y\right)^2}=\frac{x^2-y^2-xz+yz}{\left(z-x\right)\left(x-y\right)\left(y-z\right)}\left(3\right)\)
Cộng 3 vế lại với nhau ta có: \(Q=\frac{x}{\left(y-x\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
\(VP=\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}=\left(\frac{x}{y+z+t}+\frac{y+z+t}{9x}\right)+\left(\frac{y}{z+t+x}+\frac{z+t+x}{9y}\right)+\left(\frac{z}{t+x+y}+\frac{t+x+y}{9z}\right)+\left(\frac{t}{x+y+z}+\frac{x+y+z}{9t}\right)+\frac{8}{9}\left(\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\right)\)\(\ge8\sqrt[8]{\frac{x}{y+z+t}.\frac{y}{z+t+x}.\frac{z}{t+x+y}.\frac{t}{x+y+z}.\frac{y+z+t}{9x}.\frac{z+t+x}{9y}.\frac{t+x+y}{9z}.\frac{x+y+z}{9t}}+\frac{8}{9}\left(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t}\right)\)\(\ge\frac{8}{3}+\frac{8}{9}.12\sqrt[12]{\frac{y}{x}.\frac{z}{x}.\frac{t}{x}.\frac{z}{y}.\frac{t}{y}.\frac{x}{y}.\frac{t}{z}.\frac{x}{z}.\frac{y}{z}.\frac{x}{t}.\frac{y}{t}.\frac{z}{t}}=\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}=VT\left(đpcm\right)\)
Đẳng thức xảy ra khi x = y = z = t > 0
Câu 2/
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}=1\)
Điều kiện \(\hept{\begin{cases}x^2\ne0\\x^2+y^2\ne0\\x^2+y^2+z^2\ne0\end{cases}}\)
Xét \(x^2,y^2,z^2\ge1\)
Ta có: \(\hept{\begin{cases}x^2\ge1\\x^2+y^2\ge2\end{cases}}\)
\(\Rightarrow x^2\left(x^2+y^2\right)\ge2\)
\(\Rightarrow\frac{1}{x^2\left(x^2+y^2\right)}\le\frac{1}{2}\left(1\right)\)
Tương tự ta có: \(\hept{\begin{cases}\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}\le\frac{1}{6}\left(2\right)\\\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{3}\left(3\right)\end{cases}}\)
Cộng (1), (2), (3) vế theo vế ta được
\(\frac{1}{x^2\left(x^2+y^2\right)}+\frac{1}{\left(x^2+y^2\right)\left(x^2+y^2+z^2\right)}+\frac{1}{x^2\left(x^2+y^2+z^2\right)}\le\frac{1}{2}+\frac{1}{6}+\frac{1}{3}=1\)
Dấu = xảy ra khi \(x^2=y^2=z^2=1\)
\(\Rightarrow\left(x,y,z\right)=?\)
Xét \(\hept{\begin{cases}x^2\ge1\\y^2=z^2=0\end{cases}}\) thì ta có
\(\frac{1}{x^4}+\frac{1}{x^4}+\frac{1}{x^4}=1\)
\(\Leftrightarrow x^4=3\left(l\right)\)
Tương tự cho 2 trường hợp còn lại: \(\hept{\begin{cases}x^2,y^2\ge1\\z^2=0\end{cases}}\) và \(\hept{\begin{cases}x^2,z^2\ge1\\y^2=0\end{cases}}\)
Bài 2/
Ta có: \(\frac{x}{y}+\frac{y}{z}+\frac{z}{t}+\frac{t}{x}\ge4\sqrt[4]{\frac{x}{y}.\frac{y}{z}.\frac{z}{t}.\frac{t}{x}}=4>3\)
Vậy phương trình không có nghiệm nguyên dương.
Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)
=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)
\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
Như vậy:
\(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)
\(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)
\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)
\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)
\(+xy\left(z+1\right)\left(x-y\right)\)
\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)
\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\Rightarrow S=\frac{1}{xyz}\)
Đặt \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}=k\)
Áp dụng TC DTSBN ta có : \(k=\frac{x+y+z}{y+z+t}\) \(\Rightarrow k^3=\left(\frac{x+y+z}{y+z+t}\right)^3\)(1)
Ta cx có : \(k^3=\left(\frac{x}{y}\right)^3=\frac{x}{y}.\frac{x}{y}.\frac{x}{y}=\frac{x}{y}.\frac{y}{z}.\frac{z}{t}=\frac{x}{t}\) (2)
Từ (1) ; (2) \(\Rightarrow k^3=\left(\frac{x+y+z}{y+z+t}\right)^3=\frac{x}{t}\) (đpcm)