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Let $a,b,c >0$ and $q >1$. Then $ \text{if} \;\;a^q \leqslant b^q + c^q \;\;\text{then}\;\; a \leqslant b+c. $ How can I prove this?

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    substitute $b^q+c^q \le (b+c)^q$2012-06-17

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Suppose to the contrary that $a \gt b+c$. Then $a^q \gt (b+c)^q \gt b^q+c^q$.

To prove that $(b+c)^q \gt b^q+c^q$, consider the function $f(x)=(b+x)^q-b^q-x^q$. At $x=0$, $f(x)=0$. Now calculate $f'(x)$. Because $q\gt 1$, one easily shows that $f'(x)\ge 0$ if $x \gt 0$. So $f$ is an increasing function of $x$, and therefore $f(c)\gt 0$.