For any positive real $ a, b, c$, show that $(a+b+c)^\alpha<a^\alpha+b^\alpha+c^\alpha, 0<\alpha<1$.
I can show that it works for special cases like $\alpha=1/p, p\in\mathbb{N}$... but I don't know how to generalize further...
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How would you show it for the special cases? Why doesn't it generalize? – Calvin Lin Oct 11 '13 at 14:49
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I would show by raising both sides to the $p$ power and using the binomial theorem on the right. – Max Prf Oct 12 '13 at 16:03
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Calvin Lin's answer seems sufficient. However, here are a few extra notes:
It suffices just to show $(a + b)^\alpha < a^\alpha + b^\alpha$.
That would in turn follow from $(1 + x)^\alpha < 1 + x^\alpha$ for $1 \geq x > 0$.
Or in other words that $\frac{(1+x)^\alpha - 1}{x} < x^{\alpha-1}$. By the mean value theorem, the left side is of the form $\alpha c^{\alpha-1}$ for some $c$ strictly between $1$ and $1+x$. We have $\alpha c^{\alpha-1} < c^{\alpha-1} < x^{\alpha-1}$, since $\alpha-1 < 0$.
user43208
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Since you used the tag of real-analysis, I will use calculus.
Hint: Show that we have equality when $ a=b=c=0$.
Hint: Differentiate both sides with respect to each variable. Show that
$$ \alpha ( a + b + c )^{\alpha-1} \leq \alpha a^{\alpha -1 },$$
where equality holds if and only if $ b=c=0$.
Calvin Lin
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