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).



Conditions under which inequality holds for all triplets of non-negative reals
*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. 


