# Is it possible to prove $ac+bd\geq \frac{1}{2}$ if $(a+c)(b+d)=1$?

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 $$c = d = 0$$

• I'm so sorry, but I forgot to mention that $a,b,c,d$ have to be positive reals. I'll go add it right now. Commented Jul 8, 2021 at 23:25
• @BananaBlitzCoding Then try $\,a=b=0.9\,$ and $\,c=d=0.1\,$. Or, in fact, any $\,a,b\in (0,1) \setminus \{\frac{1}{2}\}$ and $\,c=1-a, \,d=1-b\,$.
– dxiv
Commented Jul 8, 2021 at 23:55

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.

• There are lots of counter examples with positive a,b,c,d as dxiv mentions. You can’t say that an inequality is true if only you ignore all the places it is false...
– Eric
Commented Jul 9, 2021 at 0:12
• Probably true ... oh well, senility sets in, it's a beautiful thing. Just wait for it!
– user77970
Commented Aug 11, 2021 at 18:17