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Following on from an earlier question, and in search of a conceptual insight, I asked myself:

Given real numbers $a \ge b \ge c \ge d \ge 0 \tag{1}$ Prove that

$a^n - b^n + c^n - d^n \ge (a - b + c - d)^n \text{ for all } \underline{n \in \mathbb{R}} \tag{2}$

First, I wrote a toy program to do a quick test, and it hinted that the conjecture may indeed be true as long as $n \ge 1 \tag{3}$

Next, I proceeded to prove it for $n \in \mathbb{N}$ by using a simple substitution technique given in this answer. $\begin{aligned}x = a-b && y = b-c && z = c-d\end{aligned} \tag{S}$

Transforms $(1)$ into: $x, y, z, d \ge 0 \tag{4}$

Transforms $(2)$ into: $(x+y+z+d)^n - (y+z+d)^n + (z+d)^n - d^n \ge (x+z)^n \tag{5}$

which is true since:

$\displaystyle \begin{aligned} & (x+y+z+d)^n - (y+z+d)^n + (z+d)^n - d^n - (x+z)^n \\ = & \sum_{r=1}^n \binom{n}{r}(x+z)^{n-r}(y+d)^r - \sum_{r=1}^n \binom{n}{r}z^{n-r}(y+d)^r + \sum_{r=1}^{n-1}d^{n-r}z^r \\= &\sum_{r=1}^n \binom{n}{r}(y+d)^r\left((x+z)^{n-r} - z^{n-r} \right)+ \sum_{r=1}^{n-1}d^{n-r}z^r \\= & \sum_{r=1}^n \binom{n}{r}(y+d)^r\left(\sum_{q=1}^{n-r} \binom{n-r}{q}x^q z^{n-r-q}\right)+ \sum_{r=1}^{n-1}d^{n-r}z^r \\ =& \text{sum of non-negative terms}\\\ge & 0\end{aligned}$

And this is where I got stuck, as I don't know how to prove it for any real $n \ge 1$.

I would appreciate any help in this direction.

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Anant
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1 Answers1

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As you already noted, the problem is equivalent to proving $$(a+b+c+d)^\alpha-(b+c+d)^\alpha+(c+d)^\alpha-d^\alpha\ge(a+c)^\alpha$$ for $a,b,c,d\ge0$.

Lemma: If $\alpha\ge1$, $x\ge y\ge0$ and $r\ge0$, then $(x+r)^\alpha-(y+r)^\alpha\ge x^\alpha-y^\alpha$.

Proof: It suffices to show that $(x+r)^\alpha-x^\alpha$ is an increasing function in $x$. Taking the derivative we have that $\alpha(x+r)^{\alpha-1}\!-\alpha x^{\alpha-1}=\alpha((x+r)^{\alpha-1}\!-x^{\alpha-1})\ge0$.

Therefore $$(a+c+(d+b))^\alpha\!-\!(c+(d+b))^\alpha+(c+d)^\alpha\!-\!(0+d)^\alpha\ge(a+c)^\alpha\!-\!c^\alpha+c^\alpha\!-\!0^\alpha=(a+c)^\alpha$$

user2345215
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  • Wow, that's a beautiful solution! Thank you :) – Anant Apr 01 '14 at 20:01
  • If possible, can you explain how you arrived at the insight for this problem; I always struggle to get one :) – Anant Apr 01 '14 at 20:10
  • @Anant I started by guessing lower estimates. So I thought what if I eliminated some common letters from $(a+b+c+d)^\alpha-(b+c+d)^\alpha$. It's quite intuitive you actually can do that, because $x^\alpha$ in convex for $\alpha\ge1$, so it should be obvious the difference of function values between two points with a fixed difference increases as you move them further on the $x$-axis. That's exactly what the lemma says. Now it's not hard to try to eliminate some letter to get the required bound. – user2345215 Apr 01 '14 at 20:18
  • Thanks for the explanation! Very instructive. – Anant Apr 01 '14 at 20:24