41 Replies
@Apu
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For 23, the function x²/x-1 is minimum at 2, so that should do it
Since p,q,r,s,t are independent
how did u get that?
Oh wait you're in class 9 or below
in first ques apply AM GM
yeah
what???
but am gm on a^2, b^2, c^2 will give a sol but that wont satisfy a + b + c = 0
tried it
any other way of applying?
(x+1)+(1/x-1) works for x²/x-1 I think
Yeah
$$x+1+\frac{1}{x-1}$$
Opt
oh got it
Now AM-GM
this is greater than eq 2 for all x
ok
4 no?
Not 2
hmm
right mb
$$2+(x-1)+\frac{1}{x-1}$$
Opt
so the ans will be 20 for 23?
and 22?
Trying, wait
thx
Do you have the answer key?
I'm getting 26 but idk
I didn't use the a+b+c=0 condition for it
dont think you can do that
yeah
is anyone here
Oh wait I got something I think
22. You'll get (ab +bc+ca)=-1/2, then you can use am-gm
If I'm not wrong
AM-GM works only on positive numbers though no?
Yeahh rightt
I was thinking only how could it be this easy
Nah the idea is right. Square the equation to get a^2b^2+b^2c^2+c^2a^2=1/4 and then apply am-gm
Since a^2, b^2 and c^2 are positive.
If not zero.
We could directly apply it then using a^2 + b^2 + c^2 = 1, why give a+b+c=0
guys it restricts the possibilities
@coolguy. check for equality
Hmm what equality
Okay I get this i didn't read it properly you squared it again
Kewl
this too shd directly give the result upon applying am gm
Again as Monushrules said, ut restricts the possibility. I think you get 2 inequalities and the one by a²+b²+c² is broader than required
Rightt makes sense
We need to narrow it down
Kewl
Got your point now brdr
Thank you brdr fr explanation 😭
@YesArnavPie123
Bhai jyada phaal gaye isme
Just let c be negative since one of them has to be negative
We can assume the other two are positive since if you negate all the variables it would still be a solution
Then you can write -c=(a+b)=d
Now the conditions become
A^2+b²+ab=1/2, a²b²(a+b)²>=?
Expand (a+b)² and substitute the value,also you can use amgm to get ab is less than or equal to 1/6
Than subsitute
idk
pretty sure we cant use am gm
as min value will lead to a^2 = b^2 = c^2
so |a| = |b| = |c|
but then u have to use the fact a + b + c = 0
$$x+1+\frac{1}{x-1}$$
Brownie
