Functions

[bright hornet]

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[real tinsel]

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[arctic quail]

try using AM-GM

@bright hornet

take $\frac{16^{x^2 + y} + 16^{x+y^2}}{2}$ as AM

[strong forum]

wolfqz

[bright hornet]

Hi

[bright hornet]

AM>=GM?

[arctic quail]

yeah

[bright hornet]

$\frac{16^{x^2 + y} + 16^{x+y^2}}{2} \ge \sqrt{16^{x^2 + y}×16^{x+y^2}}$

[strong forum]

nahihopayega

[arctic quail]

yee

[bright hornet]

Hmmmm

[arctic quail]

so

$\frac{1}{2} \geq \sqrt{16^{x^2+y+x+y^2}}$

[strong forum]

wolfqz

[bright hornet]

Half?

[arctic quail]

cuz the numerator there is equal to 1

given

[bright hornet]

Oh right

[arctic quail]

we can represent this as

$2^{-1} \geq 2^{2x^2 + 2x+2y^2+2y}$

[strong forum]

wolfqz

[bright hornet]

Yes

[arctic quail]

$-1 \geq 2\left(x^2+x+y^2+y \right)$

[strong forum]

wolfqz

[arctic quail]

$\frac{-1}{2} \geq x^2+x+y^2+y$

[strong forum]

wolfqz

[arctic quail]

hmm

$x^2+x+y^2+y+\frac{1}{2} \leq 0$

[strong forum]

wolfqz

[arctic quail]

@nahin

@bright hornet yahaan tak samjha

i mean

do u understand till here

hmm we can treat this like 2 quadratics and get a sum of factors of each maybe

let me try

[bright hornet]

Yeah

[arctic quail]

right

so

we only have 1 constant term so we make 2 constant terms out of it by writing 1/2 = 1/4+1/4

$x^2+x+\frac{1}{4} + y^2+y+\frac{1}{4} \leq 0$

$\left(x+\frac{1}{2} \right)^2 + \left(y+\frac{1}{2}\right)^2 \leq 0$

[strong forum]

wolfqz

[bright hornet]

[arctic quail]

that's what i mean

by splitting it into 2 quadratics

one in x and one in y

[bright hornet]

That's circle with no radius

[bright hornet]

Got it. Didn't know before

[arctic quail]

abandon all graphing for a moment

[arctic quail]

yeah,cuz we got 2 variables we preferrably want them in different equations so i made 1 quadratic for y and one for x

Now logic time!!

i have this

$\left(x+\frac{1}{2} \right)^2 + \left(y+\frac{1}{2}\right)^2 \leq 0$

[strong forum]

wolfqz

[arctic quail]

but the question said REAL ordered pairs

so we are only considering reals

$\left(x+\frac{1}{2} \right)^2 \in \bR \setminus \bR^- \forall x \in \bR$

because squaring makes it positive

@bright hornet

oh wait is this notation there in your class?

it just means that it is a positive real number, for all real values of x

[bright hornet]

Yeah got it

[arctic quail]

same goes for y

$\left(y+\frac{1}{2} \right)^2 \in \bR \setminus \bR^- \forall y \in \bR$

wait actually this isn't exactly the correct notation

[strong forum]

wolfqz

wolfqz

[arctic quail]

this means its a non negative real

because a square can be 0 also

and the only non negative real that is actually lesser than or equal to 0 is 0 itself...

so we convert the inequation into an equation

[bright hornet]

Hah. That set needs better notation

[arctic quail]

$\left( x+ \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 = 0$

[strong forum]

wolfqz

[arctic quail]

eh its fine, it cant rly get better lol

[bright hornet]

Alright

[arctic quail]

do u understand how i did this

[bright hornet]

Yeah I should have realised that here but realised that way too late

So 1 ordered pair. .

[arctic quail]

yeah

each of them equal to 0

by same logic

so 1 pair

[bright hornet]

Yeah

Thank you @arctic quail

[feral pagoda]

@bright hornet has given 1 rep to @arctic quail

[arctic quail]

no problem <3

u can type +close now if ur done

[bright hornet]

Alr

+close