1. ## Inequality

hello,
i have to proof $a^p+b^p \le(a^2+b^2)^{\frac{p}{2}}$for not negative real numbers and 2 $\le p<\infty$. I dont know how can i do that. can you help me? i really want to learn it
redrose5

2. ## Re: Inequality

Use the Binomial Theorem on the right hand side

3. ## Re: Inequality

ok, thank you=).
$(a^p+b^p)^2\le ((a^2+b^2)^{\frac{p}{2}})^2 \rightsquigarrow(a^p+b^p)^2=a^{2p}}+2a^pb^p+b^{2p} \le ((a^2+b^2)^{\frac{p}{2}})^2=(a^2+b^2)^p=\sum_{k=0} ^{p}\binom{p}{k}(a^2)^{p-k}(b^2)^{k}=a^{2p}+pa^{2p-2}b^2+\frac{p(p-1)}{2}a^{2p-4}b^4+\frac{p(p-1)(p-2)}{6}a^{2p-6}b^6+....+b^{2p}$
is it correct? but why is
$2a^pb^p\le pa^{2p-2}b^2+\frac{p(p-1)}{2}a^{2p-4}b^4+\frac{p(p-1)(p-2)}{6}a^{2p-6}b^6+....}$ correct?

4. ## Re: Inequality

The inequality is clearly true if $a=b$ or if $p=2.$

So, suppose that $a>b$ and $p>2,$ then we require

$a^{p}+b^{p}\leq (a^{2}+b^{2})^{p/2},$

$a^{p}\left(1+(\frac{b}{a})^{p}\right)\leq \left[a^{2}\left(1+(\frac{b}{a})^{2}\right)\right]^{p/2},$

$a^{p}\left(1+(\frac{b}{a})^{p}\right)\leq a^{p}\left(1+(\frac{b}{a})^{2}\right)^{p/2},$

and, if we let $b/a=x,$

$1+x^{p}\leq (1+x^{2})^{p/2}, \quad x<1.$

Now, if $p>2$ and $x<1,$ then $x^{p} so $1+x^{p}<1+x^{2}$ and therefore

$1+x^{p} < (1+x^{2})^{p/2}$, as required.