If $c>0$ and the equation $3a{x}^2+4bx+c=0$  has no real root, then

The given equation is $3a{x}^2+4bx+c=0$  ….. ( 1 )

Given that Eq.( 1 ) has no real root

$\therefore$   $b^2-4ac<0$

$\Rightarrow$  $ac>\frac{4b^2}{3}>0$         $\left(\because c>0\right)$

$\therefore$                   $a>0$

If  $f\left(x\right)=3a{x}^2+4bx+c$

$\therefore$   $f\left(x\right)>0$

or   $3a{x}^2+4bx+c>0,\forall x\in R$

$\therefore$   $3a{\left(-1\right)}^2+4b\left(-1\right)+c>0$

$\Rightarrow$  $3a+c>4b$