Inequality $\lambda{f(\lambda)-f(s)\over s- \lambda} - f(s)[1-f(\lambda)]\ge 0$ under some conditions

Question

Let $f(s)=\int_0^\infty e^{-sx}b(x)dx$, $s>0$. Here $b(x) \ge 0$ is the probability density function, $\int_0^\infty b(x)dx=1$, with finite mean $\bar{b}=\int_0^\infty xb(x)dx$. It is known that

1. $b(x)$ is such, that $f(s)\ge {1 \over 1+s \bar{b}}$ for $s>0$, and
2. $0<\lambda \bar{b} <1$, and
3. $0.5<f(\lambda)<1$.

I need to prove that for any $\lambda>0$ and $s>0$, which satisfy conditions (1)-(3), it holds that $$ \lambda{f(\lambda)-f(s)\over s- \lambda} - f(s)[1-f(\lambda)]\ge 0. $$
which can also be written as: $$ {f(s)[f(\lambda)-1] \over \lambda} \ge {f(s)-f(\lambda)\over s- \lambda}. $$

I came across this problem, while comparing stationary distributions of unfinished work in two single-server queues. From the physics of the queues I am convinced that the above inequality holds. Tried everything I knew and could find (cauchy-shwarz and holder inequalities, inequalities for convex functions etc.), but can't prove.

Condition (1) is important. It is satisfied by distributions $b(x)$ of class Harmonic New Worse Than Used In Expectation (see page 3 here and on page 4 one can find examples of specific disrtibutions which satisfy (1)).

I would apreciate any useful hint and advice.

Answer 1

${\color{blue}{\textrm{Edit: This answer is wrong since the inequality is not true. }}}$
See: Prove or disprove that $(1+\mathrm{e}s)\int_0^\infty \frac{\mathrm{e}^{-sx}}{2x\sqrt{2\pi}} \mathrm{e}^{-\frac{(\ln x + 1)^2}{8}}\mathrm{d} x \ge 1$

An example (Please check it)

A lognormal distribution: $b(x) = \frac{1}{2x\sqrt{2\pi}} \mathrm{e}^{-\frac{(\ln x + 1)^2}{8}},\ x > 0$.

$\bar{b} = \int_0^\infty x b(x) \mathrm{d} x = \mathrm{e}$. Let $\lambda = \frac{1}{5}$ and $s = \frac{1}{2}$

i) Check condition 1): Is it true that $$(1+\mathrm{e}s)\int_0^\infty \mathrm{e}^{-sx}\frac{1}{2x\sqrt{2\pi}} \mathrm{e}^{-\frac{(\ln x + 1)^2}{8}}\mathrm{d} x \ge 1, \ \forall s > 0?$$ I can not prove it. It appears true according to numerical evidence.

ii) Check condition 2): True.

iii) Check condition 3): True.

However, $\lambda \frac{f(\lambda) - f(s)}{s - \lambda} - f(s)(1 - f(\lambda)) < 0$.