Đề có gì đó sai sai

1: \(a^2-4b^2-2a-4b\)

\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)

\(=\left(a+2b\right)\left(a-2b-2\right)\)

2: \(x^3+2x^2-2x-1\)

\(=\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2+3x+1\right)\)

Đặt \(P=a+b+c\)

\(P^2=\left(a+b+c\right)^2=\left(1.a+\dfrac{1}{2}.2b+\dfrac{1}{3}.3c\right)^2\le\left(1^2+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2\right)\left(a^2+4b^2+9c^2\right)\)

\(\Rightarrow P^2\le\dfrac{49}{36}\left(a^2+4b^2+9c^2\right)=\dfrac{49}{36}\)

\(\Rightarrow-\dfrac{7}{6}\le P\le\dfrac{7}{6}\)

\(P_{min}=-\dfrac{7}{6}\) khi \(\left(a;b;c\right)=\left(-\dfrac{6}{7};-\dfrac{3}{14};-\dfrac{2}{21}\right)\)

\(P_{max}=\dfrac{7}{6}\) khi \(\left(a;b;c\right)=\left(\dfrac{6}{7};\dfrac{3}{14};\dfrac{2}{21}\right)\)

\(\Leftrightarrow\left(a+c\right)^2+b^2+2b\left(a+c\right)+\left(a+c\right)^2+b^2-2b\left(a+c\right)>4b^2\)

\(\Leftrightarrow\left(a+c\right)^2>b^2\)

\(\Leftrightarrow a+c>b\) (luôn đúng theo BĐT tam giác)

Vậy BĐT đã cho được chứng minh

\(\dfrac{a+b}{3}=\dfrac{b+c}{5}=\dfrac{c+a}{6}\\ \Leftrightarrow\left\{{}\begin{matrix}5a+5b=3b+3c\\5c+5a=6b+6c\\6a+6b=3c+3a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a+2b-3c=0\left(1\right)\\5a-6b-c=0\left(2\right)\\a+2b-c=0\left(3\right)\end{matrix}\right.\)

Từ \(\left(1\right)\left(2\right)\Leftrightarrow8b-4c=0\Leftrightarrow2b=c\)

Từ \(\left(1\right)\left(3\right)\Leftrightarrow4a-4c=0\Leftrightarrow a-c=0\Leftrightarrow a=c=2b\)

\(\Leftrightarrow ac-4b^2=2b.2b-4b^2=4b^2-4b^2=0\left(đpcm\right)\)

\(\left(a-1\right)^2\ge0\Rightarrow a^2+1-2a\ge0\Rightarrow a^2+1\ge2a\left(1\right)\)

\(\left(2b-3\right)^2\ge0\Rightarrow4b^2+9-12b\ge0\Rightarrow4b^2+9\ge12b\left(2\right)\)

\(\left(c\sqrt[]{3}-\sqrt[]{3}\right)^2\ge0\Rightarrow3c^2+3-6c\ge0\Rightarrow3c^2+3\ge6c\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow a^2+1+4b^2+9+3c^2+3\ge2a+12b+6c\)

\(\Rightarrow a^2+4b^2+3c^2+1+9+3\ge2a+12b+6c\)

\(\Rightarrow a^2+4b^2+3c^2+13\ge2a+12b+6c\)

\(\Rightarrow a^2+4b^2+3c^2\ge2a+12b+6c-13\)

mà \(2a+12b+6c-13>2a+12b+6c-14\)

\(\Rightarrow a^2+4b^2+3c^2>2a+12b+6c-14\)

\(\Rightarrow dpcm\)

⇔(a−1)2+(2b−3)2+3(c−1)2+1>0 (luôn đúng)

$\Rightarrow$ BĐT ban đầu đúng