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Given $a$ and $b$ are non negative real number and $m\in\mathbb{R}_{\ge1}$

Can it be shown that, If $b\ge a$ then

$$\left|\sum_{i=0}^{n-1}(-1)^i(a+ib)^m\right |\le\left|\sum_{i=0}^{n}(-1)^i(a+ib)^m\right |$$

Example let $a=0$,$b=1$ so $|1^m-2^m\pm...+(-1)^{n-1}(n-1)^m|\le|1^m-2^m\pm...+(-1)^{n}(n)^m|$

I know about triangular inequality but not get exact proof. thanks.

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Hint: square both sides and try mathematical induction (treat $n$ even and odd separately), then you'll find some famous inequality useful.

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Let $c_n=(-1)^i(a+ib)^m$ and $S_n \equiv \sum_{i=0}^n c_n$.

$$ \left| S_{n-1} \right| \leqslant \left| S_n \right| \iff S_n^2-S_{n-1}^2 \geqslant 0 \iff c_n^2+2c_n S_{n-1}\geqslant 0. $$

Case 1: $n$ is even, $n=2k$, $c_n\geqslant 0$. $$ 2 S_{n-1} +c_n = 2\sum_{i=0}^{2k-1}c_i + c_n = c_0 + \sum_{i=0}^{k-1} (c_{2i}+2c_{2i+1}+c_{2i+2}). $$

Apply Jensen's inequality, $$ c_{2i} + 2c_{2i+1} + c_{2i+2} = (a+2ib)^m-2[a+2(i+1)b]^m + [a+2(i+2)b]^m \geqslant 0. $$

Therefore $c_n^2+2c_n S_{n-1}=c_n (2S_{n-1} + c_n) \geqslant 0.$

Case 2: $n$ is odd, $n=2k+1, c_n \leqslant 0.$ $$ 2 S_{n-1} +c_n = 2\sum_{i=0}^{2k}c_i + c_n = 2c_0 +c_1 + \sum_{i=1}^k (c_{2i-1}+2c_{2i}+c_{2i+1}). $$

Again, Jensen's inequality gives $$ c_{2i-1} + 2c_{2i} + c_{2i+1} = -[a+(2i-1)b]^m+2(a+2ib)^m - [a+(2i+1)b]^m \leqslant 0. $$

And, $2c_0+c_1=2a^m-(a+b)^m \leqslant (a+b)^m-(a+b)^m = 0$.

Therefore $c_n^2+2c_n S_{n-1}=c_n (2S_{n-1} + c_n) \geqslant 0$ also holds for odd $n. \blacksquare$

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