Is an exponential function strictly increasing?

Question

Let $a, b$ and $c$ are nonnegative real numbers such that $a \geq b+c$, then I want to show that $a^r \geq b^r + c^r$ for all $r \geq 1$.

For this I need to show that for $r \geq 1 $, the exponential function $p(r) = a^r-b^r-c^r$ is strictly increasing.
Showing that the derivative is positive is not working.

Kindly help. Any little help or suggestion or tip on how can I proceed will be really appreciated. Thanks in Advance.

Answer 1

If $x,y>0$, and $1\ge x+y$, then $1\ge x^r+y^r$ for $r\ge 1$.

Let $f:[1,\infty)\to\mathbb R$ defined by $f(r)=x^r+y^r-1$. Since $x,y>0$ and $x+y\le 1$ it follows $x,y\in(0,1)$. Then $r\mapsto x^r$ and $r\mapsto y^r$ are decreasing maps, and therefore $f$ is decreasing, too. Since $f(1)\le 0$ we get $f(r)\le 0$ for all $r\ge 1$.

Now set $x=b/a$ and $y=c/a$. (If $a=0$, then $b=c=0$ and there is nothing to prove. The cases $b=0$ or $c=0$ are also obvious.)

Answer 2

Hint It is enough to prove that $$\left(1+x\right)^r \geq 1+x^r $$ which is easy since the derivative is positive on $(0, \infty)$.

Once you get this, setting $x=\frac{b}{a}$ gives you $$(a+b)^r \geq a^r+b^r$$ and you know that $$c^r \geq (a+b)^r$$