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Complex number inequalities
jhz   1
N 5 minutes ago by MS_asdfgzxcvb
Source: 2025 CTST P18
Find the smallest real number $M$ such that there exist four complex numbers $a,b,c,d$ with $|a|=|b|=|c|=|d|=1$, and for any complex number $z$, if $|z| = 1$, then\[|az^3+bz^2+cz+d|\le M.\]
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Double factorial inequality
Snoop76   3
N Yesterday at 8:02 AM by Snoop76
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Show that: $$2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}>\sum_{k=0}^n (2k+1)!!{n\choose k}$$Note: consider $(-1)!!=1$ and $n>1$
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Snoop76
Feb 7, 2025
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Double factorial inequality
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#1 h  Feb 7, 2025, 9:02 PM
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Show that: $$2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}>\sum_{k=0}^n (2k+1)!!{n\choose k}$$Note: consider $(-1)!!=1$ and $n>1$
This post has been edited 1 time. Last edited by Snoop76, Feb 7, 2025, 9:15 PM
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#6 h  Feb 25, 2025, 12:34 PM
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#9 h  Mar 23, 2025, 6:44 AM
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#10 h  Yesterday at 8:02 AM • 1 Y
Y by MihaiT
We note: $a_n=\sum_{k=0}^n (2k-1)!!{n\choose k}$, $b_n=\sum_{k=0}^n (2k+1)!!{n\choose k}$ and $c_n=2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}-\sum_{k=0}^n (2k+1)!!{n\choose k}$

Using Pascal identity $a_{n+1}=\sum_{k=0}^{n+1} (2k-1)!!{n+1\choose k}=\sum_{k=0}^{n+1} (2k-1)!!{n\choose k}+\sum_{k=0}^{n+1} (2k-1)!!{n\choose k-1}$ or

$a_{n+1}=a_n+b_n$

$b_{n+1}-a_{n+1}=\sum_{k=0}^{n+1} (2k+1)!!{n+1\choose k}-\sum_{k=0}^{n+1} (2k-1)!!{n+1\choose k}=\sum_{k=0}^{n+1}2k(2k-1)!!{n+1\choose k}$ or

$b_{n+1}-a_{n+1}=2(n+1)\sum_{k=0}^{n+1} (2k-1)!!{n\choose k-1}=2(n+1)b_n$

We have: $c_{n+1}-c_n=2(n+1)a_{n+1}-b_{n+1}-2na_n+b_n=2n(a_n+b_n)-2(n+1)b_n+a_n+b_n-2na_n+b_n$ or

$c_{n+1}-c_n=a_n>0$, $a_0=1$ results with simple induction that $c_n>0$ for $n>1$.
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