Sole Constraint

[It's late so I might be wrong with my signs. Please verify and see if this is correct.]

Cross multiplying the terms and simplifying, we obtain $2a^2 b^2 c^2 = 1 - a^2b^2 - b^2c^2 - c^2a^2$ .

We will apply titu's lemma creatively, and force the $ab+bc+ca$ term to appear on the "correct" side:

$2 = \frac{ b^2c^2 }{ a^2b^2c^2 + b^2c^2 } + \frac{ c^2a^2 } { a^2b^2c^2 + c^2 a^2 } + \frac{ a^2 b^ 2 } { a^2b^2c^2 + a^2b^2 } \geq \frac{ ( bc+ca+ab) ^2 } { 3a^2b^2c^2 + b^2c^2 + c^2a^2 + a^2b^2 }$

Cross multiplying, we get

$6 a^2b^2c^2 + 2 (b^2c^2 + c^2a^2 + a^2b^2) \geq (ab+bc+ca)^2$

Now, substituting in $2a^2b^2c^2 = 1 - a^2b^2 - b^2c^2 - c^2a^2$ , we get

$3 \geq (ab+bc+ca)^2 + a^2b^2 + b^2c^2 + c^2a^2$

Finally, since $a^2b^2 + b^2c^2 + c^2a^2 \geq \frac{1}{3} (ab+bc+ca)^2$ , we can conclude that

$3 \geq \frac{4}{3} (ab+bc+ca)^2 \Rightarrow \frac{3}{2} \geq (ab+bc+ca)$

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Note: Almost every inequality that we apply naively gives the wrong direction. IE It is true that $a^2 + b^2 + c^2 \geq \frac{3}{2}$ and $a + b + c \geq \frac{ 3 \sqrt{2} } { 2 }$ , but clearly neither of that will help us to maximize $ab + bc + ca$ .

My initial path for this solution originally focused on $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ ab+bc+ca}{abc}$ , which is what led me to looking at $\frac{ ( \frac{1}{a} ) ^2 } { (\frac{1}{a} )^2 + 1 }$ , which then led to $\frac{ b^2 c^2 } { a^2b^2c^2 + b^2c^2 }$ .

Another way of using Titu's lemma. First, we change the condition into $\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}=1$ Then we have $1=\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\stackrel{Titu's Lemma}\geq\frac{(a+b+c)^2}{a^2+b^2+c^2+3}$ $\Leftrightarrow a^2+b^2+c^2+3\geq a^2+b^2+c^2+2(ab+bc+ca)$ $\therefore ab+bc+ca\leq\frac{3}{2}$ The equality holds when $a=b=c=\frac{1}{\sqrt{2}}$