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If $n \in \mathbb{N}$ with $n \geq 2$ and $a,b \in \mathbb{R}$ with $a+b >0$ and $a \neq b$, then $$2^{n-1}(a^n+b^n)>(a+b)^n.$$

I tried to do it with induction. The induction basis was no problem but I got stuck in the induction step: $n \to n+1$

$2^n(a^{n+1}+b^{n+1})>(a+b)^{n+1} $ $ \Leftrightarrow 2^n(a\cdot a^n + b\cdot b^n)>(a+b)(a+b)^n$ $\Leftrightarrow a(2a)^n+ b(2b)^n>(a+b)(a+b)^n$ dont know what to do now :/

Michael Hardy
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Weenzeal
  • 33

3 Answers3

5

You can write that as $$\frac{{{a^n} + {b^n}}}{2} > {\left( {\frac{{a + b}}{2}} \right)^n}$$

Think convexity.

Pedro
  • 122,002
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Note that $a\not=b$ implies $(a-b)(a^n-b^n)\gt0$. Using this and the inductive hypothesis, we get

$$\begin{align} (a+b)^{n+1}&=(a+b)^n(a+b)\cr &\lt2^{n-1}(a^n+b^n)(a+b)\cr &=2^{n-1}(a^{n+1}+ab^n+a^nb+b^{n+1})\cr &=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a^{n+1}-ab^n-a^nb+b^{n+1})\cr &=2^n(a^{n+1}+b^{n+1})-2^{n-1}(a-b)(a^n-b^n)\cr &\lt2^n(a^{n+1}+b^{n+1}) \end{align}$$

Barry Cipra
  • 79,832
1

Start from the other side:

$$(a+b)^{n+1} < (a+b)(2^{n-1}(a^n+b^n))= 2^{n-1}a^{n+1}+2^{n-1}a^{n}b+2^{n-1}ab^{n}+2^{n-1}b^{n+1}$$

Now, prove that $(a^{n}-b^{n})(a-b)>0$

which implies

$$2^{n-1}a^{n}b+2^{n-1}ab^{n}< 2^{n-1}a^{n+1}+2^{n-1}b^{n+1} \,.$$

P.S. If you are familiar, the last inequality: $$a^{n}b+ab^{n}< a^{n+1}+b^{n+1} $$

also follows immediately from the AM-GM inequality.

N. S.
  • 132,525
  • There's a typo in the "Now prove" line: the plus sign should be a minus sign. – Barry Cipra Nov 07 '13 at 00:22
  • There are also typos in the first displayed equation: the exponents of all the final $a$'s and $b$'s should increase from $n-1$ and $n$ to $n$ and $n+1$ (which will necessitate subsequent changes as well). – Barry Cipra Nov 07 '13 at 00:27
  • @BarryCipra Ty, fixed the first type. The second one is no typo there, $P(n)$ implies that inequality. If you combine that one with the last inequality I get in the proof, you get exactly $P(n+1)$.. Note that I am starting from the LHS of $P(n+1)$ and going towards the RHS. It is a direct proof, not a check that $P(n+1)$ holds. – N. S. Nov 07 '13 at 01:26
  • N.S., take another look at the first line. If you ignore the $2^{n-1}$'s, you've written $(a+b)(a^n+b^n)=a^n+a^{n-1}b+ab^{n-1}+b^n$. (Actually you also omitted the exponent from the final $b$, but the real problem is that the total degree on the right hand side should be $n+1$, not $n$.) – Barry Cipra Nov 07 '13 at 01:39
  • @BarryCipra Ty, fixed... I should go to sleep, too many mistakes :) – N. S. Nov 07 '13 at 01:43