If $a+b+c<0$ and $ax^2+bx+c=0$ has no real roots, is it true that $c$ must be less than $0$? I decided to look at the graphs of the parabola to solve this problem. These are only two types that will fit this problem: parabolas that "open" upwards and parabolas that "open" downwards. These parabolas would never touch the $x$-axis, and this rules out the upwards opening parabola because $a+b+c>0$. However, I am not sure how to continue from here.  

By the quadratic formula a parabola has roots at $x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$ if such is a real number.  
The only way for there not to be roots is if $b^2 - 4ac < 0$.  Or if $b^2 < 4ac$.
If $c > 0$ then for this to happen we must have $0 < b^2 <4ac$ so $a > 0$ as well.
But $a+b+c < 0$ so $b$ must be negative.  "More negative than $a,c$ are positive".  In other words $b < -(a+c)<0$ or $|b| > |a+c|$.
So $b^2 > (a+c)^2 = a^2 + 2ac + c^2$
Now by AM-GM $a^2 + c^2 \ge 2 \sqrt{a^2c^2} = 2ac$
So $b^2 > a^2 + 2ac + c^2 > 4ac$ and thus real roots exist.
...
To get a grasp geometrically.
$ax^2 + bx + c = a(x^2 + \frac ba x) + c= a(x^2 + \frac ba x + \frac {b^2}{4a^2}) + c -  \frac {b^2}{4a^2}) = a(x + \frac b{2a})^2 + c -  \frac {b^2}{4a})$.
$a$ tells whether the parabola "goes up" or "goes down". $c -  \frac {b^2}{4a})$ is the y value of the tip of the parabola.
to Not have roots either the tip is above the $x$-axis and it points up ($c > \frac {b^2}{4a}; a > 0$.  Or the tip is below the $x$-axis and it points down ($c < \frac {b^2}{4a}; a < 0$).
In second case if $a < 0$ then $c<  \frac {b^2}{4a} < 0$.
If the first case $c >0$ and $4ac > b^2$.  ... which puts us where we were above.
