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Initially I had to show that for $0\le n \le 1$ and $a,b > 0$:

$$a^n + b^n \ge (a+b)^n$$

I made $u(x) = a^x + b^x - (a+b)^x$

$$u'(x) = a^x \log a + b^x \log b - (a+b)^x \log (a+b)$$

Now I get stuck. Cannot find where to go. please help me.

gilly
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2 Answers2

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This is a simple inequality once you see that it is homogeneous in $a, b$. Dividing both sides by $(a+b) ^{n} $ we can see that it reduces the problem to the case when $a+b=1$. And then we have $0<a<1,0<b<1$. Since $0\leq n\leq 1$ it follows that $a^{n} \geq a, b^{n} \geq b$ and hence $$a^{n} +b^{n} \geq a+b=1=(a+b)^{n}$$ as desired.

Paramanand Singh
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Note this is equivalent to $$1 + \left(\frac{b}{a}\right)^n \geq \left(1 + \frac{b}{a}\right)^n$$ so it suffices to show $1 + t^x \geq (1+t)^x$ for $x \in [0, 1]$ and $t \geq 0$.

Consider $$f(t) = 1+t^x - (1+t)^x,$$ so $$f'(t) = x t^{x-1} - x(1+t)^{x-1}=x(t^{x-1}-(1+t)^{x-1}).$$

But, $$t < 1+t \Rightarrow (1+t)^{x-1} \leq t^{x-1},$$ since $x \leq 1$.

So, $f'(t) \geq 0$ and $f(t)$ is an increasing function. But note $f(0) = 0$, so $f(t) \geq 0$ for all $t \geq0$.

Matheus Manzatto
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B. Mehta
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  • This is what i plan to do. Thank you, where do I make mistake – gilly Nov 01 '17 at 17:57
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    @gilly You used the wrong variable. You took the exponent $x$ to be the variable whereas in this answer we use the base as variable, which leads to a less complicated derivative which is easier to analyze. – Jack M Nov 01 '17 at 17:57
  • @JackM thank you alot! – gilly Nov 01 '17 at 18:03