Prove $(a+2b+3c+4d)a^ab^bc^cd^d < 1$

[anemone]

The real numbers $a,\,b,\,c,\,d$ are such that $a\ge b \ge c \ge d>0$ and $a+b+c+d=1$.

Prove that $(a+2b+3c+4d)a^ab^bc^cd^d<1$.

 

[jvicens]

Thank you for bringing up this interesting problem. I would like to provide a proof for the statement you have presented.

First, let us rewrite the expression as $(a^a b^b c^c d^d)(a+2b+3c+4d)$. We can see that the first part of the expression, $(a^a b^b c^c d^d)$, is always less than or equal to $1$ since $a,b,c,d$ are all positive and their sum is $1$. This is due to the fact that the geometric mean is always less than or equal to the arithmetic mean.

Now, let us focus on the second part of the expression, $(a+2b+3c+4d)$. We can rewrite this as $(a+b)+(b+c)+(c+d)+d$. Since $a\ge b \ge c \ge d$, we know that $a+b \ge b+c \ge c+d \ge d$. Therefore, by the rearrangement inequality, we have $(a+b)+(b+c)+(c+d)+d \ge (a+b)+(b+c)+(c+d)+d$. This means that the second part of the expression is always greater than or equal to $1$.

Combining the two parts, we have $(a^a b^b c^c d^d)(a+2b+3c+4d) \le 1 \cdot 1 = 1$. Therefore, we can conclude that $(a+2b+3c+4d)a^ab^bc^cd^d < 1$.

I hope this explanation helps in understanding the problem and its solution. If you have any further questions or concerns, please do not hesitate to ask.