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

Solution 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.

Solution 2:

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.

Solution 3:

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 $(x+y)^{n+1} \leq 2^n(x^{n+1} + y^{n+1})$