\(Tacó:\)
\(\left\{{}\begin{matrix}\left|2x+1\right|\ge0\\\left|3x+2\right|\ge0\\\left|4x+3\right|\ge0\end{matrix}\right.\Rightarrow\left|2x+1\right|+\left|3x+2\right|+\left|4x+3\right|\ge0\Rightarrow x-1\ge0\Rightarrow x\ge1\Rightarrow\left\{{}\begin{matrix}2x+1>0\\3x+2>0\\4x+3>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|2x+1\right|=2x+1\\\left|3x+2\right|=3x+2\\\left|4x+3\right|=4x+3\end{matrix}\right.\Rightarrow2x+1+3x+2+4x+3=x-1\Leftrightarrow9x+6=x-1\Leftrightarrow8x=-7\left(\text{vô lí}\right)\)
\(Vậy:x\in\varnothing\)
\(2,\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\Leftrightarrow\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2\ge\left(ax\right)^2+2axby+\left(by\right)^2\Leftrightarrow\left(ay\right)^2+\left(bx\right)^2\ge2axby\Leftrightarrow\left(ay\right)^2-2axby+\left(bx\right)^2\ge0\Leftrightarrow\left(ay-bx\right)^2\ge0\left(\text{luôn đúng}\right).\text{Vậy BĐT đã được chứng minh}\)
