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# Prove connection between power mean and arithmetic mean of a function

Let be a continuous, strictly monotonic function on with inverse , and let be given positive real numbers. Then define, This is called the mean of with respect to . (When for , this coincides with the th power mean from this exercise).

Show that Proof. Since is the inverse of we know for all in the range of , i.e., for all such that there is some such that .

By the definition of then, we have that So, if is in the domain of then we are done. Since is the inverse of it’s domain is equal to the range of . We show that this value is in the range of using the intermediate value theorem.

Without loss of generality, assume is strictly increasing (the alternative assumption, that is strictly decreasing will produce an almost identical argument). Then, since are all positive real numbers we have . (Here if we’d assumed that was strictly decreasing the roles inequalities would be reversed.) Then we have, Hence, by the intermediate value theorem, since there must be some such that Thus, is in the domain of , so # Use the pth power mean to prove an inequality

If for , prove that Proof. For not all equal, we may apply the previous exercise with and with , to obtain, Further, if , then Thus, the requested inequality holds for any , and as a bonus, we prove equality holds if and only if # Prove the pth power mean is less than the (2p)th power mean

We recall the definition of the th power mean .

For , and with , we define the th power-mean as: Now, for , prove for not all equal.

Proof. From the Cauchy-Schwarz inequality we know that for real numbers and , we have with equality if and only if there is some such that for all . Letting and we have This inequality is strict since if equality held there would exist some such that for all , but this would imply for all , contradicting our assumption that the are not all equal. Since (see here), this implies 