Proof of the inequality $(x+y)^n\leq 2^{n-1}(x^n+y^n)$

Can you help me to prove that
$$(x+y)^n\leq 2^{n-1}(x^n+y^n)$$
for $n\ge1$ and $x,y\ge0$.

I tried by induction, but I didn’t get a result.

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Look. I also have tried to do it by induction. It is obvious that it holds for $n=1$ and $n=2$. Assume that it also holds for $n$. Let’s prove that inequality for $n+1$:
$$
2^n(x^{n+1} + y^{n+1}) – (x+y)^{n+1} = 2^{n}(x^{n+1} + y^{n+1}) -(x+y)^n(x+y)\geq 2^n(x^{n+1} + y^{n+1}) – 2^{n-1}(x^n + y^n)(x+y) =
2^{n-1}(x^{n+1} + y^{n+1} – x^ny – y^nx) = 2^{n-1}(y(y^n – x^n) + x(x^n – y^n))=
2^{n-1}(y^n-x^n)(y-x) = 2^{n-1}(y-x)^2(y^{n-1}+xy^{n-2}+x^2y^{n-2}+\dots+y^2x^{n-2} + x^{n-1}) \geq 0.
$$
So we have proved that $2^n(x^{n+1} + y^{n+1}) – (x+y)^{n+1} \geq 0$. It is similar to $(x+y)^{n+1} \leq 2^n(x^{n+1} + y^{n+1})$

Using Jensen’s Inequality

This is simply Jensen’s Inequality applied to the convex function $x^n$ for $n\ge1$ and a discrete measure. In fact, this can be generalized to any non-negative sequence $(\alpha_i)$ so that
$$
\sum_{i=1}^k\alpha_i=1
$$
to get
$$
\left(\sum_{i=1}^k\alpha_ix_i\right)^n\le\sum_{i=1}^k\alpha_ix_i^n
$$
In this particular case, $k=2$ and
$$
(\alpha_i)=\left(\frac12,\frac12\right)
$$
yields
$$
\left(\frac{x+y}{2}\right)^n\le\frac{x^n+y^n}{2}
$$
which is the same as
$$
(x+y)^n\le2^{n-1}(x^n+y^n)
$$


Using the AM-GM Inequality

Note that the AM-GM Inequality applied to $\{\overbrace{x^n,x^n,\dots,x^n}^{k\text{ copies}},\overbrace{y^n,y^n,\dots,y^n}^{n-k\text{ copies}}\}$ yields
$$
x^ky^{n-k}\le\frac{kx^n+(n-k)y^n}{n}\tag{AM-GM}
$$
Applying this to the Binomial Theorem yields
$$
\begin{align}
(x+y)^n
&=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}\\
&\le\sum_{k=0}^n\binom{n}{k}\frac{kx^n+(n-k)y^n}{n}\\
&=x^n\sum_{k=0}^n\binom{n}{k}\frac{k}{n}+y^n\sum_{k=0}^n\binom{n}{k}\frac{n-k}{n}\\
&=x^n\sum_{k=1}^n\binom{n-1}{k-1}+y^n\sum_{k=0}^{n-1}\binom{n-1}{k}\\[6pt]
&=2^{n-1}(x^n+y^n)
\end{align}
$$


Negative Values

The inequality is true if $x,y\ge0$ or $n$ is even. The proofs above were written assuming that $x,y\ge0$; however, $x^n$ is convex over all $\mathbb{R}$ when $n$ is even, so the Jensen proof works as is. Furthermore, if $n$ is even, changing the sign of $x$ and/or $y$ leaves the right side of $\text{(AM-GM)}$ alone (and positive), but might change the sign of the left side to be negative. In either case, $\text{(AM-GM)}$ holds for all $x,y\in\mathbb{R}$ when $n$ is even.

Note that the equation is equivalent to $$\left(\frac{x+y}{2}\right)^n\le \frac{x^n+y^n}2$$

And this can be proved by induction (with appropriate constraints to avoid multiplying an inequality by a negative number) provided $$\frac{(x^n+y^n)}2\frac{(x+y)}2\le\frac{x^{n+1}+y^{n+1}}2$$ and this reduces to $$x^{n+1}+y^{n+1}-xy^n-x^ny\ge 0$$ or $$x^n(x-y)-y^n(x-y)\ge0
$$

or $$(x^n-y^n)(x-y)\ge 0$$ And you can probably complete the details from there.


Note in response to comment – use the inductive hypothesis as follows:

$$\left(\frac{x+y}{2}\right)^{n+1}\le\frac{(x^n+y^n)}2\frac{(x+y)}2\le\frac{x^{n+1}+y^{n+1}}2$$

The first inequality comes from the hypothesis, the second will get us from $n$ to $n+1$ if we can prove it.

You’ve already seen approaches using estimates and induction. I want to show you another way to prove the claim that also points out where $2^{n-1}$ comes from.

Consider $g(x) = (x + y)^n – Ax^n – Ay^n$. $$g'(x) = -nAx^{n – 1} + n(x + y)^{n – 1}$$and $$g'(x_0) = 0 \Longleftrightarrow x_0 = \frac{y}{A^{1/(n – 1)} – 1}.$$
Now note that the best costant is the one that makes the minimum equal to $0$.
$$g(x_0) = 0 \Longleftrightarrow Ax_0^n + Ay^n = (x_0 + y)^n \Longleftrightarrow \frac{Ay^n}{(A^{1/(n-1)} – 1)^n} + Ay^n = \frac{A^{n/(n-1)}y^n}{(A^{1/n – 1} – 1)^n} \Longleftrightarrow Ay^n\Big(\frac{1}{(A^{1/(n – 1)} – 1)^n} + 1 – A^{n/(n – 1)}\frac{1}{(A^{1/(n – 1)^n} – 1)}\Big) \Longleftrightarrow \frac{1}{(A^{1/(n – 1)} – 1)^n} – A^{n/(n – 1)}\frac{1}{(A^{1/(n – 1)} – 1)^n} = -1 \Longleftrightarrow \frac{1}{(A^{1/(n – 1)} – 1)^n}(A^{1/(n – 1)} – 1) = 1 \Longleftrightarrow A^{1/(n – 1)} – 1 = 1 \Longleftrightarrow A = 2^{n – 1}.$$

While we already have an array of solutions with a variety of distinct approaches, I was surprised that no one used the possibly-simplest approach of maxima.

With the non-negativity assumptions $x,y\geq0$ we may use the max-properties $\max(x,y) \leq x+y$ and $\max(x,y)^n = \max(x^n,y^n), n \in \mathbb{N}$, to obtain the following linear/calculational/motivated proof.

$
\hspace{0.5cm}2^{n-1}(x^n + y^n) \\
\geq 2^{n-1} \cdot\max(x^n,y^n) \hspace{3cm}\text{—since $x^n,y^n\geq0$} \\
= \frac{1}{2} \cdot 2^n \cdot \max(x^n,y^n) \hspace{3cm}\text{ —arithmetic}\\
= \frac{1}{2} \cdot 2^n \cdot \max(x,y)^n \hspace{3cm}\text{ — since $x,y,n\geq0$}\\
= \frac{1}{2} \cdot (2 \cdot \max(x,y))^n \hspace{3cm}\text{ —powers}\\
= \frac{1}{2} \cdot (\max(x,y) +\max(x,y))^n \hspace{1.5cm}\text{ —arithmetic}\\
\geq \frac{1}{2} \cdot (x+y)^n \hspace{2.5cm}\text{—since $x,y \leq \max(x,y)$ and non-negative powers preserve order}\\
= \frac{(x+y)^n}{2} \hspace{5cm}\text{—notation}
$

Note that we have started with the complex expression with the goal of ${\bf simplifying}$ it. As a result, we have ${\bf discovered}$, rather than $\textit{verified}$, the result!

Note that the above proof shows that the result holds for any non-negative reals $x,y$ and any integer $n\geq0$.

Indeed, motivated-discovery is better than verification.

Best regards,

Moses

EDIT :: something’s gone wrong; we have an extra fraction at the end … if anyone could explain where I’ve erred, I’d be most-grateful!

$\vdots$

So we proved $(x+y)^n \leq 2^n(x^n+y^n)$ and this follows from what we had to prove $\ldots$