Does the prime sequence satisfy $p_n+p_m \le p_{n+m} < p_n p_m$? I did some experiments in SAGE and it seems like the prime sequence $p_n$ satisfies:
$$p_n+p_m \le p_{n+m} < p_n p_m$$
for $(n,m) \neq (1,1)$.
For $n=1$ the last inequality is Bertrand's postulate.
Here is some SAGE code to test this for some primes:
lp = list(primes(10000))
for n in range(len(lp)):
    for m in range(len(lp)):
        pn = lp[n]
        pm = lp[m]
        pnm = nth_prime(n+m+2)
        print pn*pm>=pnm,pnm>=pn+pm, n+1,m+1,n+m+2,pn,pm,pn*pm,pnm

Is this something known ( if it is true)? And if so, how does one prove or disprove it, are there heuristics?
 A: The left inequality is equivalent to the Second Hardy-Littlewood conjecture.
The k-tuple conjecutre, AKA First Hardy-Littlewood conjecture is the statement that the density of every Prime Constellation can be computed using a single general formula. If it is true, then there are infinitely many twin primes, also infinitely many prime tuples of form $(p, p+4)$, $(p, p+6)$, $(p, p+2, p+6, p+8)$, etc. Note that it does not imply the prime tuple $(p, p+2, p+4)$ is infinite, which is not true.
If k-tuple conjecture is true, then Second Hardy-Littlewood conjecture is not true. $\pi(3159)=446$, but there may be $447$-tuple primes spanning $3159$ integers. Such tuple is not yet discovered, but the formula in k-tuple conjecture suggests that the first such tuple is likely to be between $1.5\times10^{174}$ and $2.2\times10^{1198}$.

For right inequality, WLOG assume that $m \le n$. Then for $10 \le n$ and $3 \le m$, we can apply the inequality mentioned by @gammatester. $$p_{m+n}\le p_{2n}<2n\ln{(2n\ln{2n})}<4n{(\ln(n\ln n)-1)}<4p_n\le p_mp_n$$
For $n<10$, one can check all the cases manually. For $m=1$, it is Bertrand's Postulate. For $m=2$, it is proving $p_{n+2}<3p_n$. It is known that if $k\ge25$, then there exists a prime between $k$ and $1.2k$. It is trivial corollary that $p_{n+2}<1.2^3p_n<3p_n$ if $n>10$, especially since $p_{10}=29>25$.
