Bài 1:
\(A=\frac{x+1}{x^2+x+1}\)
\(\Leftrightarrow Ax^2+Ax+A=x+1\)
\(\Leftrightarrow Ax^2+Ax+A-x-1=0\)
\(\Leftrightarrow x^2\cdot A+x\cdot\left(A-1\right)+\left(A-1\right)=0\)
Để pt có nghiệm thì \(\Delta\ge0\)
\(\Leftrightarrow\left(A-1\right)^2-4\left(A-1\right)\cdot A\ge0\)
\(\Leftrightarrow A^2-2A+1-4A^2+4A\ge0\)
\(\Leftrightarrow-3A^2+2A+1\ge0\)
\(\Leftrightarrow\left(A-1\right)\left(3A+1\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}A-1\le0\\3A+1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}A-1\ge0\\3A+1\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\frac{-1}{3}\le A\le1\left(chon\right)\\1\le A\le\frac{-1}{3}\left(loai\right)\end{matrix}\right.\)
Vậy \(minA=\frac{-1}{3};maxA=1\)
Bài 2:
\(VT=\Sigma\frac{c\left(a+b+c\right)+ab}{a+b}=\Sigma\frac{ac+bc+c^2+ab}{a+b}=\Sigma\frac{\left(a+c\right)\left(b+c\right)}{\left(a+b\right)}\)
Áp dụng BĐT quen thuộc \(x^2+y^2+z^2\ge xy+yz+zx\) :
\(VT\ge\Sigma\sqrt{\frac{\left(a+c\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+c\right)\left(a+b\right)}}=\Sigma\sqrt{\left(b+c\right)^2}=\Sigma\left(b+c\right)=2\cdot\left(a+b+c\right)=2\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{\left(a+c\right)\left(b+c\right)}{a+b}=\frac{\left(a+c\right)\left(a+b\right)}{b+c}=\frac{\left(a+b\right)\left(b+c\right)}{a+c}\)