<s>[INST] Find minimum of the $a+c+e$ if $a+b+c+d+e=100$. 
Let $a\ge b\ge c \ge d \ge e \ge0$ and $a+b+c+d+e=100$. Find minimum of
$$a+c+e$$

My work so far:
$100=a+b+c+d+e\le 5a \Rightarrow a\ge20$
I think that $a+c+e \ge 50$ (if $a=40,b=40,c=10,d=10,e=0$ or $a=b=49, c=d=e=1$).
 [/INST] $$100=a+b+c+d+e\le a+a+c+c+e=2a+2c+2e-e.$$ Thus $$50\le 50+\frac e2\le a+c+e.$$ So, the minimum of $a+c+e$ is $50$ and is achieved if $e=0.$
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