How to prove this inequality by using induction?

Question

If $x,y$ are distinct real numbers such that $x+y>0$ and $n\ge 1$, then $2^{n-1}(x^n+y^n)\ge (x+y)^n$.

It is obvious for $n=1$. How to do the rest by using induction?

This is true for any $n \in \mathbb{R}_{\geq 1}$. See here. If you are just looking for a proof and not by induction, I would vote to close this down as duplicate of the other. — , May 14 '13 at 02:18

Prism · Answer 1 · 2013-05-14T02:52:19.290

3

Suppose the claim is true for $n$. Then, multiply $2^{n-1}(x^{n}+y^{n})\ge (x+y)^n$ by $(x+y)$ on both sides to obtain (note that here we are using $x+y>0$) $$ 2^{n-1}(x^{n}+y^{n})(x+y) \ge (x+y)^{n+1}$$ To prove the inductive hypothesis it is enough to show that $$ 2^{n}(x^{n+1}+y^{n+1}) \ge 2^{n-1}(x^{n}+y^{n})(x+y)$$ which boils down to proving (I will let you proceed the intermediate steps on your own): $$ x^{n+1}+y^{n+1}\ge x^n y + xy^{n}$$ Without loss of generality, assume $x\ge y$. Then: $$ x^{n+1}+y^{n+1}-x^{n}y-xy^{n}=(x^{n}-y^{n})(x-y)\ge 0$$ as desired.

Edit: as pointed out by Ross Millikan, we do not need the assumption that $x$ and $y$ are distinct.

edited May 14 '13 at 02:52

answered May 14 '13 at 02:23

Prism

11,782

Just wondering, how is it that you can claim $x\ne y$ and not lose generality? – apnorton May 14 '13 at 02:24
@anorton: It says in the question that $x$ and $y$ are distinct. – Prism May 14 '13 at 02:26
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@anorton: since the comparisons are $\ge$ it works fine if $x=y$. The two sides are equal then. – Ross Millikan May 14 '13 at 02:29
@anorton Ooopss: sorry I meant $x>y$. Thanks! Corrected. – Prism May 14 '13 at 02:31
@Prism: But the request was $\ge$ and, as you said, $x \neq y$ was specified. It works both ways. You can change yours to $\gt$ if you want. – Ross Millikan May 14 '13 at 02:32
@RossMillikan: Thanks. I have edited it accordingly. – Prism May 14 '13 at 02:37
How do you know that $x^n-y^n\ge 0$? – user73564 May 14 '13 at 02:44
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@user73564 Suppose $y$ were negative, then the condition $x+y>0$ implies $0<|y|<x$, so that $x^n\ge |y|^n\ge y^n$. – Karl Kroningfeld May 14 '13 at 02:51

score 1 · Answer 2 · edited May 14 '13 at 02:33

Hint: convexity of $f\colon t\in\mathbb{R}_+\mapsto t^n$.

Observe that the result is equivalent to $$ \frac{x^n+y^n}{2} \geq \left(\frac{x+y}{2}\right)^n $$ that is $$ \frac{f(x)+f(y)}{2} \geq f\!\left(\frac{x+y}{2}\right) $$

However, this method does not rely on induction.

For one with induction: assume the result is known for $n\geq 1$: $$ \frac{x^n+y^n}{2} \geq \left(\frac{x+y}{2}\right)^n $$ Then (as $x+y>0$) $$ \left(\frac{x+y}{2}\right)^{n+1} \leq \frac{x+y}{2}\cdot\frac{x^n+y^n}{2} = \frac{x^{n+1}+y^{n+1}+xy^n+yx^n}{4} $$ Remains to prove that $$ \frac{x^{n+1}+y^{n+1}+xy^n+yx^n}{4} \leq\frac{2x^{n+1}+2y^{n+1}}{4} $$ i.e. that $$ x^{n+1}+y^{n+1} - (xy^n+yx^n) \geq 0 $$ But the LHS can be rewritten $$ x^{n+1}+y^{n+1} - (xy^n+yx^n) = (x^{n}-y^{n}) (x-y) $$ which is indeed always positive as $x+y>0$ (whether $x>y$ or $y>x$, at least one of them is positive; a further analysis shows that, independently of the parity of $n$, both factors will have the same sign).

You wrote $$\frac{f(x^n)+f(y^n)}{2} $$ instead of $$\frac{f(x)+f(y)}{2} $$ — Pedro, May 14 '13 at 02:33

How to prove this inequality by using induction?

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