Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế với vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(21x^2-2xy-4y^2\right)=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{7x}{4}\\y=-3x\end{matrix}\right.\) thế xuống pt dưới:

\(\Rightarrow\left[{}\begin{matrix}1+y^2=5\\1+\left(\dfrac{7x}{4}\right)^2=5\left(1+x^2\right)\\1+9x^2=5\left(1+x^2\right)\end{matrix}\right.\) \(\Leftrightarrow...\)

Đặt \(a=\dfrac{yz}{x^2};b=\dfrac{zx}{y^2};c=\dfrac{xy}{z^2}\)

Áp dụng BĐT BSC:

\(\dfrac{1}{a^2+a+1}+\dfrac{1}{b^2+b+1}+\dfrac{1}{c^2+c+1}\)

\(=\dfrac{x^4}{x^4+x^2yz+y^2z^2}+\dfrac{y^4}{y^4+y^2zx+z^2x^2}+\dfrac{z^4}{z^4+z^2xy+x^2y^2}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)}\)

Ta cần chứng minh: 

\(\dfrac{\left(x^2+y^2+z^2\right)^2}{x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)}\ge1\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\ge x^4+y^4+z^4+x^2y^2+y^2z^2+z^2x^2+xyz\left(x+y+z\right)\)

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

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

\(\Rightarrow dpcm\)

Đẳng thức xảy ra khi \(a=b=c=1\)

\(x^2y^2+xy+1=x^2\)

\(\Leftrightarrow4x^2y^2+4xy+4=4x^2\)

\(\Leftrightarrow\left(2xy+1\right)^2+3=4x^2\)

\(\Leftrightarrow\left(2x-2xy-1\right)\left(2x+2xy+1\right)=3=1.3=\left(-1\right).\left(-3\right)\)

TH1: \(\left\{{}\begin{matrix}2x-2xy-1=1\\2x+2xy+1=3\end{matrix}\right.\Leftrightarrow...\)

TH2: \(\left\{{}\begin{matrix}2x-2xy-1=3\\2x+2xy+1=1\end{matrix}\right.\Leftrightarrow...\)

TH3: \(\left\{{}\begin{matrix}2x-2xy-1=-1\\2x+2xy+1=-3\end{matrix}\right.\Leftrightarrow...\)

TH4: \(\left\{{}\begin{matrix}2x-2xy-1=-3\\2x+2xy+1=-1\end{matrix}\right.\Leftrightarrow...\)