Inequality for a decreasing, concave function on [0,1] Suppose I have a function $f:[0,1]\rightarrow \mathbb{R}_{\geq0}$ that is decreasing and concave. Let $a,b,c,d\in(0,1)$, with $a>b$ and $a+b>1$. I am trying to prove the following inequality:
$$ 
f\Bigg(\frac{(1-a)\cdot c}{(1-a)\cdot c + b\cdot d }\Bigg)\cdot \big( (1-a)\cdot c + b\cdot d \big)
+ 
f\Bigg(\frac{a\cdot c}{a\cdot c + (1-b)\cdot d }\Bigg)\cdot \big( a\cdot c + (1-b)\cdot d \big) \leq \\
f\Bigg(\frac{(1-b)\cdot c}{(1-b)\cdot c + a\cdot d }\Bigg)\cdot \big( (1-b)\cdot c + a\cdot d \big)
+ 
f\Bigg(\frac{b\cdot c}{b\cdot c + (1-a)\cdot d }\Bigg)\cdot \big( b\cdot c + (1-a)\cdot d \big) 
$$
My approach was the following. Since $f$ is also subadditive (it's concave and $f(0)\geq0$), you can show that
$$
f\Bigg(\frac{(1-a)\cdot c}{(1-a)\cdot c + b\cdot d }\Bigg)\cdot \big( (1-a)\cdot c + b\cdot d \big) \leq f\big( (1-a)\cdot c \big), 
$$ and
$$
f\Bigg(\frac{a\cdot c}{a\cdot c + (1-b)\cdot d }\Bigg)\cdot \big( a\cdot c + (1-b)\cdot d \big) \leq f\big(a\cdot c \big),
$$
after which I got stuck.. Does anyone know how to proceed, or perhaps know a better way to approach this?
EDIT: Similarly, (using the subadditivity), you can also show that
$$
f\Bigg(\frac{(1-b)\cdot c}{(1-b)\cdot c + a\cdot d }\Bigg)\cdot \big( (1-b)\cdot c + a\cdot d \big)\geq f(1) \cdot (1-b)\cdot c,
$$
and
$$
f\Bigg(\frac{b\cdot c}{b\cdot c + (1-a)\cdot d }\Bigg)\cdot \big( b\cdot c + (1-a)\cdot d \big) \geq f(1) \cdot b\cdot c.
$$
If you combine these inequalities you will get the sufficient condition:
$$f\big( (1-a)\cdot c \big) + f\big(a\cdot c \big) \leq f(1)\cdot c,$$
which is not satisfied because of the decreasing property of $f$. So this condition is too "loose", probably because the inequalities are too loose..

A counterexample:
Let $f(x) = 2 - x^2$.
Let $a = 2/3, ~ b = 1/2, ~ c = 1/3, ~ d = 3/4$.
We have
$$\mathrm{LHS} = \frac{35}{72}f(8/35) + \frac{43}{72}f(16/43) = \frac{18589}{9030}$$
and
$$\mathrm{RHS} = \frac{2}{3}f(1/4) + \frac{5}{12}f(2/5) = \frac{247}{120}.$$
We have $\mathrm{LHS} - \mathrm{RHS} = \frac{3}{12040}$.
