$\log_{2} \frac{1-ab}{a + b} = 2ab + a + b -3$. Find min of $(a +b)$

Question

Problem: $a$, $b$ are positive real numbers and $\log_{2} \frac{1-ab}{a + b} = 2ab + a + b -3$. Find the minimum value of $P = a + b$

I reached this point and had no idea how to proceed. Please advise: $2^{2(1-ab) + 1}(1 - ab) = 2^{a + b}(a + b)$

Answer 1

Let us take notations

$$S=a+b, \ \ P=ab$$

keeping in mind the classical inequality (Arith. Mean vs. Geom. Mean)

$$S \ge 2\sqrt{P}\tag{1}$$

With these notations, your relationship, which can be written :

$$8(1-ab)=2^{a+b+2ab}(a+b)$$

becomes :

$$8(1-P)=2^{S+2P}S$$

(1) implies :

$$8(1-P) \ge 2^{2\sqrt{P}+2P}2 \sqrt{P}$$

$$\underbrace{4(1-P)}_{f(P)} \ge \underbrace{4^{\sqrt{P}+P}\sqrt{P}}_{g(P)}\tag{2}$$

$f$ is decreasing with $f(0)=4,f(1)=0$, $g$ is increasing with $g(0)=0,g(1)=16$ as can be seen on the following graphical representation

Therefore their curves have a unique intersection point.

In fact the abscissa $P_0$ of this point is the unique positive root $P_0$ of equation :

$$P+\sqrt{P}=1\tag{3}$$

(check it by plugging the LHS of (3) in (2)).

Equation (3) above can be solved explicitly because, setting $p=\sqrt{P}$, it becomes a quadratic equation, giving :

$$P_0=\left(\tfrac12(\sqrt{5}-1)\right)^2=\tfrac12(3-\sqrt{5}) \approx 0.381966\tag{4}$$

In this way, the maximal value of $P$ is $P_0$.

From there, the minimal value of $S$ is obtained by using the result saying that for a fixed value of the sum of two numbers, their product is maximal when they are equal.

Indeed, consider a fixed value of S :

• For example if $S=2$, is it possible to have a corresponding P ? The answer is no because $P$ is maximum when $a=b$ ; in this case we would have $a=b=1$ giving $P=1$ which is impossible.

• Can we have $S=1$ ? Now, the answer is yes because it suffices to take $a=b=\tfrac12$ giving $P=ab=\frac14$ which is possible, etc.

One has understood with these examples that the optimal choice leading to a minimal $P$ is to take :

$$a=b=\tfrac12S \ \text{in such a way that} \ P:=ab=P_0$$ which means $$a=\sqrt{P_0} \ \iff \ S=2a=2\sqrt{P_0}$$

I.e., the optimal/minimal $S$ is (see (4)) :

$$S=S_0=\sqrt{5}-1$$