TITLE: If $a^2=b^2+c^2$ and $0<n<2$ prove $a^n<b^n+c^n$
QUESTION [1 upvotes]: If $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, prove
(a) if $n>2$ then $a^n>b^n+c^n$,
(b) if $0<n<2$ then $a^n<b^n+c^n$.
Part (a) was easy to prove: $a^2=b^2+c^2$ and $a,b,c$ are positive real numbers, so $a>b$ and $a>c$. Then O can show: $b^n+c^n=b^2(b^{n-2})+c^2(c^{n-2})<b^2(a^{n-2})+c^2(a^{n-2})=a^{n-2}(b^2+c^2)=a^{n-2}\times a^2=a^n \implies b^n+c^n<a^n$
But I stuck in part (b). Can I prove it in a way I proved part (a)?

REPLY [1 votes]: Like you did in (a), we can do as fellows: when $0<n<2$
$$a^n=\frac{a^2}{a^{2-n}}=\frac{b^2}{a^{2-n}}+\frac{c^2}{a^{2-n}}<\frac{b^2}{b^{2-n}}+\frac{c^2}{c^{2-n}}=b^n+c^n.$$
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