Mình nghĩ là thế này
Ta có: x2+1>0 ∀xϵR
x2+2x+3=(x+1)2+1>0 ∀xϵR
x2+4x+5=(x+2)2+1 >0 ∀xϵR
nên \(\sqrt{x^2+1}+2\sqrt{x^2+2x+3}\ge3\sqrt{x^2+4x+5}\)
\(\Leftrightarrow\sqrt{x^2+1}+2\sqrt{\left(x+1\right)^2+1}\ge3\sqrt{\left(x+2\right)^2+1}\)
\(\Leftrightarrow x+1+2\left(x+1\right)+2\ge3\left(x+2\right)+3\)
\(\Leftrightarrow x+3+2x+2\ge3x+6+3\)
\(\Leftrightarrow3x+5\ge3x+9\Leftrightarrow0x\ge4\) (vô nghiệm)
Vậy S=∅
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x^2+2x+3}=b>0\end{matrix}\right.\)
\(a+2b\ge3\sqrt{2b^2-a^2}\)
\(\Leftrightarrow a^2+4b^2+4ab\ge18b^2-9a^2\)
\(\Leftrightarrow5a^2+2ab-7b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(5a+7b\right)\ge0\)
\(\Leftrightarrow a-b\ge0\) (do \(5a+7b>0\))
\(\Leftrightarrow a\ge b\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x^2+2x+3}\)
\(\Leftrightarrow x^2+1\ge x^2+2x+3\Leftrightarrow x\le-1\)
Vậy nghiệm của BPT là \(x\le-1\)
