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[snow forge]

Help

[high tangle]

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[fleet zenith]

The answer is not a single number

For example, $(a,b,c)$ might be $(1,0,0)$, and $ab+bc+ca = 0$. If $(a,b,c) = (\frac 1 {\sqrt 3},\frac 1 {\sqrt 3},\frac 1 {\sqrt 3)}$ then, $ab+bc+ca$ becomes $1$ instead

[ruby rock]

k12byda5h

[celest pumice]

all we can pull outta this alone is $\sqrt{1+2(ab+bc+ca)} = a+b+c$
there are infinite values to $a+b+c$ so infinite values to $ab+bc+ca$

[ruby rock]

yoavmal isobar

[old crystal]

hmmm

let's take ab+bc+ca as x here

now, u know $(a+b+c)^2 = a^2 + b^2 + c^2+ 2ab + 2bc + 2ca$

[ruby rock]

No Lifer?

[old crystal]

$(a+b+c)^2 = 1 + 2x$

[ruby rock]

No Lifer?

[old crystal]

$a + b + c = \sqrt{1+2x}$

[ruby rock]

No Lifer?

[old crystal]

hmmm

hey wait...

bruh

Alr @snow forge here's a step by step solution
This requires the AM GM inequality, so if you aren't familiar with that I suggest you go search it up

[celest pumice]

@old crystal brother, AM GM inequality cannot do this, it literally has infinite values

[old crystal]

as we know:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

and $(a+b+c)^2 \geq 0$ for any real a,b,c

Now, it's given that $a^2 + b^2 + c^2 = 1$

So, $1 + 2ab + 2bc + 2ca \geq 0$

$ab + bc + ca \geq \frac{-1}{2}$

Now, AM GM inequality states that

$a+b \geq 2\sqrt{ab}$

lets just plug in $a = a^2, b = b^2 ; a = b^2, b = c^2 ; a = c^2, b = a^2$ to get 3 vital equations:

$a^2 + b^2 \geq 2ab$

$b^2 + c^2 \geq 2bc$

$c^2 + a^2 \geq 2ca$

adding these things, we get:

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

$1 \geq ab + bc + ca$

we also have $ab+bc+ca \geq \frac{-1}{2}$

So:

$ab+bc+ca \in [\frac{-1}{2},1]$

The above result is all the possible values of ab+bc+ca

Hope this helps!

@snow forge

[ruby rock]

No Lifer?

[obsidian stirrup]

can we also do it using GM>= HM?

[old crystal]

I encourage you to try, but I feel like AM GM is simpler
It could work though!

If it works, post your solution here and ping the OP

[obsidian stirrup]

okay thanks!

[celest pumice]

he has = sign, this still doesnt solve it, but its all u can get

because its unsolvable

[old crystal]

yeah
It does help to decrease it to a range of values though