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?
$ \text{if} \;\;a^q \leqslant b^q + c^q \;\;\text{then}\;\; a \leqslant b+c. $
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calculus
inequality
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1substitute $b^q+c^q \le (b+c)^q$ – 2012-06-17
1 Answers
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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 $f(c)\gt 0$.