For consecutive primes $a\lt b\lt c$, prove that $a+b\ge c$.

Yes, this is true. In 1952, Nagura proved that for $n \geq 25$, there is always a prime between $n$ and $(6/5)n$. Thus, let $p_k$ be a prime at least $25$. Then $p_k+p_{k+1} > 2p_k$. But by Nagura's result we have that $p_{k+2} \leq 36/25 p_k < 2p_k$. Finally, one can easily check by hand that the result holds for small primes.

Ramanujan (1919), see Eq. (18):

$$\pi(x) - \pi(x/2) \ge 2 \quad \text{ for } x\ge 11 $$

Whence, with $x= 2p_k$ for $p_k \ge 7$, $$p_{k+2} \le 2 p_k \lt p_k+p_{k+1}, $$ and $5\le 2+3$, $7\le 3+ 5$, $11 \le 5+7$.

As a matter of fact, P. L. Chebyshev knew already that for any $\epsilon > \frac{1}{5}$, there exists an $n(\epsilon) \in \mathbb{N}$ such that for all $n\geq n(\epsilon),$

$\pi((1+\epsilon)n)-\pi(n)>0.$

In [2], one can find a short report on the problem of determining the smallest $n(\epsilon)$ explicitly once that $\epsilon$ has been fixed.

References

[1] P. L. Chebyshev. Mémoire sur les nombres premiers. Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.

[2] H. Harborth & A. Kemnitz. Calculations for Bertrand's Postulate. Mathematics Magazine, 54 (1), pp. 33-34.