$1.{3^{x + 1}} - {2^{2x + 1}} - {12^{\frac{x}{2}}} < 0\Leftrightarrow 3.3^x-2.4^x-(\sqrt{12})^x<0$
$\Leftrightarrow 3-2(\frac{4}{3})^x-(\frac{\sqrt{12}}{3})^x<0\Leftrightarrow 3-2(\frac{4}{3})^x-(\sqrt{\frac{4}{3}})^x<0$
Đặt $t=(\sqrt{\frac{4}{3}})^x>0$
$BPT\Leftrightarrow
3-t-2t^2<0\Leftrightarrow t>1$ ( do $t>0$)$\Leftrightarrow$$x>0$
$2.$ Có : $a^2-2a+1\geq 0,\forall x\Rightarrow a^2\geq a+a-1$
$\Rightarrow a^2+b^2+c^2\geq (a+b+c)+(a+b+c)-3=a+b+c$(do giả thiết $a+b+c=3$)
$\Rightarrow a^2+b^2+c^2\geq 3$
Tương tự ta có $a^4+b^4+c^4\geq a^2+b^2+c^2+(a^2+b^2+c^2)-3$
$\Rightarrow a^4+b^4+c^4\geq a^2+b^2+c^2 $
$\Rightarrow 2(a^4+b^4+c^4)\geq a^2+a^4+b^2+b^4+c^2 +c^4\geq 2a^3+2b^3+2c^3$
(Theo Cauchy)
$\Rightarrow a^4+b^4+c^4\geq a^3+b^3+c^3 $ Đẳng thức xảy ra $\Leftrightarrow a=b=c=1$