$-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$. Show that $-3\cdot7^x+2\cdot6^x+2\cdot5^x-2\cdot4^x+3^x>0$, for $-1\le x<0$.
Here its plotting.
By first derivative it involves natural log and expression become more difficult.

Note that $a^x$ is a decreasing convex function of $a>0$ for $x<0$. Hence
$$-3\cdot7^x+2\cdot 6^x+5^x>0$$
$$5^x-2\cdot 4^x+3^x>0$$
Add these up to get the needed inequality.
