请教一道不等式证明

数学小黄

设x > 0, y > 0,0<α<β,证明
$$(x^{\alpha }+y^{\alpha })^{\frac{1}{\alpha }}>(x^{\beta }+y^{\beta })^{\frac{1}{\beta }}$$

realnumber

设$f(t)=\frac{\ln{(x^t+y^t)}}{t}.t\ge 0$,以下求导数，证明它是减函数.
\[f'(t)=\frac{\frac{x^t\ln{x}+y^t\ln{y}}{x^t+y^t}t-\ln{(x^t+y^t)}}{t^2}\le 0\]
等价于证明下式\[tx^t\ln{x}+ty^t\ln{y}\le (x^t+y^t)\ln{(x^t+y^t)}\]
而以下两式显然成立$tx^t\ln{x}\le x^t\ln{(x^t+y^t)},ty^t\ln{y}\le y^t\ln{(x^t+y^t)}$
所以$f(t)$是t的减函数.
更多的x,y,z,...类似可以证明.

kuing

\begin{align*}
(x^{\alpha }+y^{\alpha })^{\frac{1}{\alpha }}>(x^{\beta }+y^{\beta })^{\frac{1}{\beta }}&\iff x^{\alpha }+y^{\alpha }>(x^{\beta }+y^{\beta })^{\frac{\alpha }{\beta }} \\
& \iff\left( \frac{x^{\beta }}{x^{\beta }+y^{\beta }} \right)^{\frac{\alpha }{\beta }}+\left( \frac{y^{\beta }}{x^{\beta }+y^{\beta }} \right)^{\frac{\alpha }{\beta }}>1,
\end{align*}
显然成立

realnumber

恩，这样简单~~