<s>[INST] $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero. 
Let $a,b,c$ and $d$ be real numbers such that $a^4+b^4+c^4+d^4=16$. Then $a^5+b^5+c^5+d^5=32$ if and only if one of $a,b,c,d$ is $2$ and others are zero.

Why does this hold?
 [/INST] Suppose $a,b,c,d\geq 0$ and set $x=a^4,\,y=b^4,\,z=c^4,\,w=d^4$. Then
$$f(x,y,z,w)=x^{5/4}+y^{5/4}+z^{5/4}+w^{5/4}$$
is strictly convex and thus restricted to the domain
$$\{x,y,z,w\geq 0: x+y+z+w=16\}$$
it attains the maximum only in the extremal points, which are the four vertices $(16,0,0,0)$ and cyclicals.
If one number is less then zero than taking the moduli the value of $a^5+b^5+c^5+d^5$ strictly increases.
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