Conditions under which inequality holds for all triplets of non-negative reals I have the following inequality:
$$ \left(\frac{1}{3}  
   \left(
   \left(\frac{a+b}{2}\right)^3 +   
   \left(\frac{a+c}{2}\right)^3 +
   \left(\frac{b+c}{2}\right)^3     \right)   \right)^\frac{1}{3}
   \leq
   \left(\frac{a^p+b^p+c^p}{3}\right)^\frac{1}{p}$$
which is true $\forall a,b,c \in \mathbb{R}^+\cup\{0\} $.
I would like to show that this inequality above holds if and only if $ p \geq \frac{3}{2} $.
i.e. the inequality holds for all non-negative reals $a,b,c $ as long as $ p \geq \frac{3}{2}$.
Any help whatsoever would be appreciated, so please comment even if you don't have a full solution (even proving 1 direction of the implication would be very helpful).
P.S. I'm not sure exactly what to tag this, I've tagged it as convex-optimisation for now, as one of the things I've tried unsuccessfully is maximising/minimising one the sides subject to the constraint of the other side being held constant. If this is tagged wrong, please let me know the right tag(s).
 A: it would be quite a nice result, but unfortunately it's proven quite difficult.
I'm not sure about "nice" (after all, one can invent infinitely many inequalities for 3 positive numbers) but it is certainly not difficult.
WLOG, $a+b+c=3$.
Write $a=1+A, b=1+B, c=1+C$ with $A,B,C\ge -1, A+B+C=0$. 
Then the LHS equals
$$
\left[\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3\right]^{1/3}
$$
Part 1: $p$ must be $\ge 3/2$
Consider the case when $A,B,C\to 0$ and look at both sides up to the second order. We have
$$
LHS=1+\frac{A^2+B^2+C^2}{12}+\text{higher order terms}\,;
\\
RHS=1+\frac{p-1}6(A^2+B^2+C^2)+\text{higher order terms}
$$
whence $p-1\ge \frac 12$ is necessary.
Part 2: $p=3/2$ works (and, thereby, any larger $p$ works as well by Holder).
We need to prove that 
$$
\sqrt{\frac{(1-\frac A2)^3+(1-\frac B2)^3+(1-\frac C2)^3}3}
\le \frac{(1+A)^{3/2}+(1+B)^{3/2}+(1+C)^{3/2}}3
$$
Opening the parentheses, recalling that $A+B+C=0$, and using the inequality $\sqrt{1+X}\le 1+\frac X2$, we estimate the LHS from above by
$$
1+\frac{A^2+B^2+C^2}8-\frac{A^3+B^3+C^3}{48}
$$
Thus, it would suffice to show that
$$
(1+A)^{3/2}\ge 1+\frac 32A+\frac 38A^2-\frac 1{16}A^3
$$
However the RHS here is just the cubic Taylor polynomial of $A\mapsto (1+A)^{3/2}$ at $A=0$ and the fourth derivative is positive.  
