<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] We have
$$a^4\le a^4+b^4+c^4+d^4=16\implies a\le 2$$
So, we can have $a^4(a-2)\le 0$, i.e.
$$a^5\le 2a^4$$
Similarly, 
$$b^5\le 2b^4,\quad c^5\le 2c^4,\quad d^5\le 2d^4$$
giving
$$a^5+b^5+c^5+d^5\le 2(a^4+b^4+c^4+d^4)=32$$
Note here that 
$$a^5+b^5+c^5+d^5=2(a^4+b^4+c^4+d^4)$$
holds when
$$a^4(a-2)=b^4(b-2)=c^4(c-2)=d^4(d-2)=0$$
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