<s>[INST] If $a^2=b^2+c^2$ and $0If $a^2=b^2+c^2$ 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)?
 [/INST] We need to show: $\left(\dfrac{b^2}{a^2}\right)^{x} + \left(\dfrac{c^2}{a^2}\right)^{x} > 1$ for $0< x = \dfrac{n}{2}< 1$, and $a^2 = b^2 + c^2$. Let $u = \dfrac{b^2}{a^2}$, and $v = \dfrac{c^2}{a^2}$, then $0 < u,v < 1$, and $u+v = 1$, and we now prove: $u^x + v^x > 1$.
Consider $f(x) = u^x + v^x$ on $(0,1)$. We have: $f'(x) = u^x\cdot \ln u + v^x\cdot \ln v < 0$ because $\ln u < 0$, and $\ln v < 0$ for $0 < u,v < 1$. Thus $f(x)$ decreases, and $f(x) > f(1) = u+v = 1$.
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