If I prove $ac+bd\geq \frac{1}{2},$ then I solve another inequality that I have been trying to solve. This one seems simple enough, but I have already spent 30 min on it. I was wondering if it is even possible to prove this using the condition given or should I just give up on this approach? If it is possible, I was wondering if I could get a hint.
Edit: $a,b,c,d$ are positive reals.
Counter example: $a = b = 1$ and $c = d = 0$
If you exclude the counterexample in the accepted answer and restrict a, b, c, d > 0, I think you get the inequality holding if $(a^2 + b^2) < (a^2 - b + c) \epsilon$, for some small $\epsilon$.
I got there by letting $(a + c) = 1 + \epsilon$, so $1 = (b + d) + \frac{\epsilon}{a + c}$, wlog assume $a \geq 1/2$, $b \leq 1/2$, then use the above equations to get $ac = a(1 + \epsilon -a)$, $bd = b(1 - \frac{\epsilon}{a + c} - b)$
Multiplying these expressions through for ac and bd and adding them and rearranging gives $-a^2 -b^2 + a \epsilon - \frac{b \epsilon}{a + c} + b < a - 1/2$.
Since $b > 0$ a little rearrangement gives the expression I came up with.
I feel like there is a cleaner solution but I think the inequality is generally true if the counterexample is excluded.
