Let $a,b,c$ be positive real numbers. Then, there exist unique positive real $x,y$ such that $\frac{a+b+c}{3}$=$\frac{x+y}{2}$, and $\sqrt{\frac{a^2+b^2+c^2}{3}}$=$\sqrt{\frac{x^2+y^2}{2}}$.
Strangely, if $c=(a+b)/2$, then $\sqrt[3]{\frac{a^3+b^3+c^3}{3}}$= $\sqrt[3]{\frac{x^3+y^3}{2}}$. Wolfram Alpha confirms this last claim, but it feels like there should be some proof of it besides ugly expansion.
Question: Is there an approach to prove this last statement which provides intuition as to why it is true?
A slightly different approach: the system is equivalent to $$\left\{\begin{array}{c}x+y=\frac 2 3(a+b+c):=A\\ x^2+y^2=\frac 23(a^2+b^2+c^2):=B\end{array}\right.$$ By elimination, one clearly has $$xy=\frac 12(A^2-B):=C$$ which together with the first equation can be used to solve for $(x,y)$. Note that in general there are two solutions. To address your second problem, namely to obtain the following relation: $$x^3+y^3=\frac 23(a^3+b^3+c^3):=g(a,b,c),~{\rm if~}c=(a+b)/2,$$ consider $$x^3+y^3=(x+y)^3-3xy(x+y)=A^3-3CA:=f(a,b,c).$$ In order that $f(a,b,c)=g(a,b,c)$, consider the factorization $$f(a,b,c)-g(a,b,c)=\frac 2{27}(a+b-2c)(b+c-2a)(c+a-2b),$$ which equals zero if either of the following three conditions holds: $$c=(a+b)/2,a=(b+c)/2,~{\rm or~}b=(c+a)/2.$$