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I'm currently preparing for a French math contest, and I'm having some trouble proving an inequality. The inequality I'm trying to show is:

Problem. Show that \begin{equation} (a^n + b^n) \sum_{k=0}^{n} \binom{n}{k} \cdot a^k \cdot b^{n-k} \leq (a^{2n} + b^{2n}) \sum_{k=0}^{n} \binom{n}{k} \end{equation} Here, both $a$ and $b$ are assumed to be positive real numbers.

We can write this as:

\begin{equation} (a^n + b^n) (a+b)^{n} \leq 2^{n}(a^{2n} + b^{2n}) \end{equation}

I believe I need to prove that:

\begin{equation} a^{k+n} b^{n-k} + a^k b^{2n-k} \leq a^{2n} + b^{2n} \end{equation}

I would really appreciate it if someone could help me with this proof. Any insights or hints on how to approach this problem would be greatly appreciated. Thank you in advance for your assistance.

Best regards.

Sangchul Lee
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Henry D
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    $\sum_{k=0}^{n} \binom{n}{k} \cdot a^k \cdot b^{n-k} =(a+b)^n$ so $\sum_{k=0}^{n} \binom{n}{k}=2^n$ – Henry Oct 30 '23 at 15:20
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    Are there any restrictions on $a,b$? Because if not then the inequality would be false. – Kamal Saleh Oct 30 '23 at 15:24
  • @KamalSaleh Yes, there are restrictions on a and b. In the context of the inequality, both 'a' and 'b' are assumed to be positive real numbers. This restriction is necessary for the inequality to hold. – Henry D Oct 30 '23 at 15:45

3 Answers3

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Your guess is almost correct. By noting that

\begin{align*} (a^n + b^n) \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} &= a^n \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k} + b^n \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \\ &= \sum_{k=0}^{n} \binom{n}{k} (a^{n+k}b^{n-k} + a^{n-k}b^{n+k}), \end{align*}

it suffices to prove that

$$ a^{n+k}b^{n-k} + a^{n-k}b^{n+k} \leq a^{2n} + b^{2n} \tag{*} $$

for each $k = 0, \ldots, n$. To this end, note that for any non-negative integers $p$ and $q$,

$$ a^p b^q = \bigl[ \underbrace{a^{p+q} \cdot a^{p+q} \cdots a^{p+q}}_{p\text{ times}}\cdot\underbrace{b^{p+q} \cdot b^{p+q} \cdots b^{p+q}}_{q\text{ times}} \bigr]^{\frac{1}{p+q}} \leq \frac{p a^{p+q} + q b^{p+q}}{p+q} $$

by the AM-GM inequality.1) By applying this inequality, we get

\begin{align*} a^{n+k}b^{n-k} + a^{n-k}b^{n+k} &\leq \frac{(n+k)a^{2n} + (n-k)b^{2n}}{2n} + \frac{(n-k)a^{2n} + (n+k)b^{2n}}{2n} \\ &= a^{2n} + b^{2n} \end{align*}

as required.

1) This inequality actually holds for any real $p, q \geq 0$ so that $p +q > 0$, which can be proved using Young's inequality or Jensen's inequality. However, we don't need this for our proof.

Sangchul Lee
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  • Just wanted to say thanks for your help ! I didn't know about Jensen and Young's inequalities before, and they're really cool. I mean, Hӧlder's inequality is awesome, but these new ones open up a whole new world of understanding, especially the link to Cauchy-Schwarz. Thanks again! – Henry D Oct 31 '23 at 11:16
  • Please can you sens me the proof for your Inequality with a^p*b^p – Henry D Dec 18 '23 at 16:56
  • @HenryD, I found there is a typo in the inequality for $a^p b^q$, which does not affect the subsequent argument. I fixed it, and now the current version should be easy to prove. I already provide a proof using AM-GM when $p$ and $q$ are non-negative integers with $p+q>0$. – Sangchul Lee Dec 18 '23 at 17:07
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We have $$a^{2n} + b^{2n} - a^{n + k}b^{n-k} - a^{n-k}b^{n+k} = (a^{n + k} - b^{n + k})(a^{n - k} - b^{n - k})\ge 0.$$ (Note: The last inequality is easy. WLOG, assume that $a \ge b$. Then $a^{n+k} \ge b^{n+k}$ and $a^{n-k} \ge b^{n-k}$. We are done. By the way, I used this trick several times, e.g. here.)

River Li
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A less tricky approach is (letting $x:=\frac a{a+b}$) to prove that $$\forall x\in(0,1),\quad f(x):=2^n\left(x^{2n}+(1-x)^{2n}\right)-x^n-(1-x)^n\ge0.$$ For this, just notice that $f(1/2)=0$ and $f'$ is negative on $(0,1/2)$ and positive on $(1/2,1).$

Anne Bauval
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