How do I prove $(y+c+b)^\lambda + (x+c+b)^\lambda \geq (y+x+c)^\lambda + (c+2b)^\lambda$ for $0<=\lambda<=1$ $\forall y \geq x \geq b>0, c \geq b$ [closed]

Let $f(x) = (x+c+b)^{\lambda}$. It is enough to show that $$f(y)+f(x) \geq f(x+y-b)+f(b).$$ An equivalent way of saying this is $$f(x)-f(b) \geq f(x+y-b) - f(y)$$ and $$\frac{f(x)-f(b)}{x-b} \geq \frac{f(x+y-b) - f(y)}{x-b}$$ since $x>b$ ($x=b$ case follows if we check the original inequality.) Since $c+b > 0$, $f$ is differentiable for positive real $x$. Thus by mean value theorem, $\frac{f(x)-f(b)}{x-b} = f'(c_1)$ and $\frac{f(x+y-b) - f(y)}{x-b} = f'(c_2)$, where $c_1\in (b,x)$ and $c_2 \in (y,x+y-b)$. Note that $c_1 \leq c_2$. Since $0<\lambda <1$, $f'(x)$ is a decreasing function if $x$ is positive, therefore $f'(c_1) \geq f'(c_2)$, concluding the proof.