Find the maximum value...

Apply geometric mean formula now ((abcd )^4)^1/4 is equal to 16 on solving (abcd) is equal to 16 now take a^5 plus b^5 plus c^5 plus d^5 and apply geometric mean formula ((abcd)^5)^1/4 so on solving (abcd)^5/4 split now (abcd)^1/4 is equal to 2 from 1st eqn abcd is equal to 16 so (abcd)^1/4 is equal to (16)^1/4 which is equal to 2 now substitute value 2 in ((abcd)^1/4)^5 which is equal to (2)^5 which is equal to 32 .

Let $S = a^5+b^5+c^5+d^5$ and $C = a^4+b^4+c^4+d^4$ . Suppose we have values $a,b,c,d$ such that $C = 16$ .

Assume that two variables are non-zero and unequal, say $b > a$ . We will alter them slightly so that $a\to a+da$ and $b\to b+db$ , in such a way that $C = 16$ . That is,

$dC = 4a^3da + 4b^3db = 0\ \Rightarrow\ db = -\left(\frac{b}{a}\right)^3\ da.$

This changes the sum $S$ as follows:

$dS = 5a^4da + 5b^4db = -5a^4\left(\frac{b}{a}\right)^3db + 5b^4db = 5b^3db(b-a).$

This change is positive if we increase $b$ . Thus whenever two variables are non-zero and unequal, $S$ can be improved by making their difference even bigger. The case where $a = b = c = d$ is actually a minimum with $S = 16\sqrt 2$ .

It follows that the maximum value is reached when all but one variable are equal to zero. W.l.o.g., $a = 4, b = c = d = 0$ , given the maximum value $S = 32$ .