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Let $a, b$ be positive real numbers. Let $0 < x < 1$. Show that $(a+b)^x < a^x + b^x$.

The function $f(x) = (a+b)^x -a^x - b^x$ has $f(0) = -1$, $f(1) = 0$ and it seems to be an increasing function. But I am unable to show that it is increasing function.

Shailesh Kumar
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  • A demo is on https://www.desmos.com/calculator/0axazt8sqs – Shailesh Kumar Oct 31 '16 at 20:18
  • More generally, for any concave function $f : \mathbb{R}_{+} \rightarrow \mathbb{R}$ such that $f(0) \ge 0$, you can prove that $f(a+b) \le f(a)+f(b)$ (in your case, $f:t\mapsto t^x$ is differentiable, so you can use $f'$ decreasing to prove what you need) – charmd Oct 31 '16 at 20:37

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Here is another way:

WLOG we may normalize with $a+b = 1$, so $a, b \in (0, 1) \implies a^x > a,\quad b^x > b$, so $a^x+b^x > a+ b = 1$.

Macavity
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