$a + b = 2$ implies $a^c + b^c \ge 2$ for any real $c \ge 1$

Question

If $a, b, c$ are positive reals such that $c \ge 1$ and $a + b \ge 2$ then $a^c + b^c \ge 2$. Is there any elementary way to prove it without using calculus and advanced inequalities like Jensen's?

Note: I know how to prove it with Jensen, I was just wondering if it can be done using simple, basic, elementary theorems and inequalities.

Answer 1

By Bernoulli $$a^c=(1+a-1)^c\geq1+c(a-1),$$ $$b^c\geq1+c(b-1).$$ Id est, $$a^c+b^c\geq2+c(a+b-2)=2.$$

Answer 2

Look at $f(x) = (1 + x)^c + (1 - x)^c$. We have $f(0) = 2$, and

$$f'(x) = c(1 + x)^{c-1} - c(1 - x)^{c-1}$$

Thus $f'(x) \geq 0$ if $c \geq 1$ and $0 \leq x \leq 1$.

So for $0 \leq x \leq 1$ one has $f(x) \geq f(0) = 2$. Suppose $a + b = 2$. WLOG we can assume $b \geq a$. Then we can let $x = {b - a \over 2}$ in which case $1 + x = b$ and $1 - x = a$, giving your result.

Yeah, I know it's calculus but I think this is a natural elementary way to do it.