Proving Angles in Triangle ABC < 120° & Cos + Sin > -√3/3

[anemone]

All the angles in triangle $ABC$ are less than $120^{\circ}$. Prove that

$\dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}>-\dfrac{\sqrt{3}}{3}$.

 

[anemone]

Spoiler: Solution of other

[TIKZ]
\coordinate[label=left:A] (A) at (0,0);
\coordinate[label=right:B] (B) at (6, 0);
\coordinate[label=above:C] (C) at (2.4,3);
\coordinate[label=above: $A_1$] ($A_1$) at (7,3);
\coordinate[label=below: $B_1$] ($B_1$) at (7.2,0);
\coordinate[label=below: $C_1$] ($C_1$) at (12,2);
\draw (A) -- (B)-- (C)-- (A);
\draw (7,3) -- (7.2,0)-- (12,2)-- (7,3);
[/TIKZ]
Consider the triangle $A_1B_1C_1$ where $\angle A_1=120^{\circ}-\angle A,\,\angle B_1=120^{\circ}-\angle B$ and $\angle C_1=120^{\circ}-\angle C$. The given condition guarantees the existence of such a triangle.

Applying the triangle inequality in triangle $A_1B_1C_1$ gives $B_1C_1+C_1A_1>A_1B_1$, i.e.

$\sin A_1+\sin B_1>\sin C_1$ by applying the law of sines to triangle $A_1B_1C_1$.

It follows that

$\sin (120^{\circ}-A)+\sin (120^{\circ}-B)>\sin (120^{\circ}-C)$ or

$\dfrac{\sqrt{3}}{2}(\cos A+\cos B+\cos C)+\dfrac{1}{2}(\sin A+\sin B+\sin C)>0$.

Taking into account that $a+b>c$ implies $\sin A+\sin B-\sin C>0$, the above inequality can be rewritten as

$\dfrac{\sqrt{3}}{2}\cdot \dfrac{\cos A+\cos B+\cos C}{\sin A+\sin B+\sin C}+\dfrac{1}{2}>0$, from which the conclusion follows.