let $\{a_{n}\}_{\ge 0},n\in \Bbb N^{+},a_{n}\ge 0$ is decreasing. show that
$$\left(\sum_{k=1}^{n}(-1)^{k+1}a_{k}\right)^r\le \sum_{k=1}^{n}(-1)^{k+1}a^r_{k},(r>0)$$
My try: $$\Longleftrightarrow \left(a_{1}-a_{2}+\cdots+(-1)^{n+1}a_{n}\right)^r\le a^r_{1}-a^r_{2}+\cdots+(-1)^{n+1}a^r_{n}$$
following I want use Mathematical induction.But I can't How use $a_{n}$ is decreasing ,Thank you
Note that we need $r\geq 1$, as pointed out by experimentX.
We will prove this by (strong) induction, according to the parity of $n$. This follows directly from studying the $n=2$, $n=3$ case.
Claim: If $a, b \geq 0$, then $a^r + b^r \leq (a+b)^r$.
Proof: We treat $a$ as given, and $b$ can vary from 0 to $\infty$. This is true when $b =0$. Observe that $rb^{r-1} \leq r(a+b)^{r-1}$ (using $r\geq 1$), hence the result follows by integrating. $_\square$
If $n=2i$ is even, then assume the induction hypothesis is true for $2i-1$.
$$ a_{2i}^r + (a_1 - a_2 + \ldots + a_{2i-1} - a_{2i})^r \leq (a_1 - a_2 + \ldots + a_{2i-1})^r \leq a_1 ^r - a_2 ^r + \ldots + a_{2i-1}^r.$$
Hence, moving $a_{2i}^r$ over to the other side completes this step.
Claim: If $ x \geq a \geq b \geq 0$, then $ (x-a + b)^r \leq x^r - a^r + b^r$.
Proof: We treat $x, a$ as given and $b$ can vary from $0$ to $a$. Equality holds when $b=a$. Since $r (x-a+b)^{r-1} \geq r b^{r-1}$ (using $r\geq 1$), the result follows by integration. (Note that we since the differential is greater, hence the function is lower, in order for it to eventually be equal.) $_\square$
If $n=2k+1$ is odd, then assume the induction hypothesis is true for $2i-1$, we have
$$ (a_1 - a_2 + \ldots - a_{2i} + a_{2i+1} ) ^r \leq (a_1 - a_2 + \ldots + a_{2i-1})^r - a_{2i}^r + a_{2i+1}^r \\ \leq a_1 ^r - a_2 ^r + \ldots + a_{2i-1}^r- a_{2i}^r + a_{2i+1}^r $$
Hence this completes the induction step.
Note: It is in the odd step that we do require the assumption of non-decresing of the entire sequence. Interestingly, the even step didn't need non-decreasing.
We prove the calculus claims, using the fundamental theorem of calculus. Note that these are the standard tools of working with inequalities.
Theorem. If $f$ and $g$ are function such that $ f'(x) \geq g'(x)$ in the domain $[x_1, x_2]$ and $f(x_1) = g(x_1)$, then for all $ y \in [x_1, x_2]$, we have $ f(y) \geq g(y)$.
Proof: Since $f'(z) \geq g'(z)$, we have $\int_{x_1} ^y f'(z) \, dz \geq \int_{x_1}^y g'(z) \, dz $. Applying the fundamental theorem of calculus, we get that
$$ f(y) - f(x_1) \geq g(y) - g(x_1). $$
The statement then follows since $f(x_1) = g(x_1)$. $_\square$
Theorem. If $f$ and $g$ are function such that $ f'(x) \geq g'(x)$ in the domain $[x_1, x_2]$ and $f(x_2) = g(x_2)$, then for all $ y \in [x_1, x_2]$, we have $ f(y) \leq g(y)$.
Proof: Proof: Since $f'(z) \geq g'(z)$, we have $\int_{y} ^{x_2} f'(z) \, dz \geq \int_{y}^{z_2} g'(z) \, dz $. Applying the fundamental theorem of calculus, we get that
$$ f(x_2) - f(y) \geq g(x_2) - g(y). $$
The statement then follows since $f(x_2) = g(x_2)$. $_\square$
Note: These two theorems are actually the same. Can you find a change of variables that allow us to deduce the second theorem from the first, even though the latter seems surprising?
