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Jan 24, 2022 at 21:02 comment added Iosif Pinelis @Hans : Yes, you were right. Sorry about that. This is now fixed.
Jan 24, 2022 at 21:02 history edited Iosif Pinelis CC BY-SA 4.0
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Jan 24, 2022 at 20:56 comment added Iosif Pinelis @Hans : Yes, I have the same 1st derivative.
Jan 24, 2022 at 20:24 comment added Hans I just checked my calculation. It seems my statement is correct. As a reference, the first derivative from my calculation is $$\frac d{dt}M(p+th)=\Big(1+\frac1r\Big)\Big(\sum_i(p_i+th_i)^r\Big)^{\frac1r-1}\sum_i(p_i+th_i)^rh_i.$$
Jan 24, 2022 at 20:09 comment added Iosif Pinelis @Hans : I think there is no typo there. Can you re-check your calculation?
Jan 24, 2022 at 19:24 comment added Hans You are again right, Iosif. There may be one typo in your last expression though. It seems the factor $1+r$ ought to be outside of the left-most large parenthesis.
Jan 24, 2022 at 16:06 comment added Iosif Pinelis @Hans : Yes, the same method works for $r>1$ -- then we can also use the Cauchy--Schwarz inequality, with weights $p_i^{r-1}$.
Jan 24, 2022 at 15:46 vote accept Hans
Jan 24, 2022 at 15:46 comment added Hans Thank you, Iosif, as usual. I actually thought of this conventional direct method of proving the second derivative being positive late last night, but decided to write it up later this morning... I am glad though you came up with the same method. Sometimes we overlook the obvious... By the way, for $r>1$, we can prove it the way I have mentioned in my question, but can you prove it by the second derivative method as well? Now we have to compare the two terms of opposite signs.
Jan 24, 2022 at 13:28 history edited Iosif Pinelis CC BY-SA 4.0
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Jan 24, 2022 at 13:22 history answered Iosif Pinelis CC BY-SA 4.0