Some doubts.!

1st que :

given expression is same as coefficient of x^6 in the expansion of :
((1+x)¹⁰ - 1)⁵

Expand and get the answer...

yeah. the equality condition in AM-GM requires the summands to be equal.

Here's a method that uses only Cauchy Schwarz only

(1^2+1^2+1^2+1^2)(4x^2+4x^2+y^2+z^2) \ge (2x+2x+y+z)^2 by CS

Further (2x+2x+y+z) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right)\ge (1+1+1+1)^2 = 4^2 again by CS [u can also use AM-HM or AM-GM)

Therefore we have

4(4x^2+4x^2+y^2+z^2) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right)^2 \ge \left[ \left(2x+2x+y+z) \left(\frac{1}{2x} + \frac{1}{2x} + \frac{1}{y} + \frac{1}{z} \right) \right]^2 \ge 4^4

and hence the inequaliyty holds

othrwise i had thought abt sumthing else...
if u c carefully, in the first expression, equality holds whn 8x² = y² = z² = some k

in the second one, equality holds whn x^-1 = y^-1 = z^-1 = some constant

now 54 is the min value only if both are satisfied simultaneously, ie for the same values of x, y and z...which is clearly not possible...
so i think 64 is 100 % correct ans..

I think i found it

in nihit's solution

Second part

squaring the inequality can reverse the inequality because its value can be <1 also

so that solution will not be valid

:D

hm ! even the explanation given in solution booklet matches with the one given by prophet sir. :)

But still, why is AM-GM not working here !?

Well, I dont know if there is any other method. the answer 54 is got by applying AM-GM on the two factors. But that needs x,y and z to be positive. Also the equality condition is not satisfied, so that 54 is definitely a lower bound , but not the greatest lower bound for this expression.

hey...u welcum..but wait !...
i m still not convinced y that ans cannot b 54...
i mean, even i got that whn i workd with AM GM inequality...i cant find my mistake...

thanx nihit :)

well, it's done then.

Thanq everyone who involved in this forum and also the one's who tried these problems. Thanx once again :)

thanx a lot sir. :)

but if that's not applicable for jee syllabus, how to solve such problems within jee level.?

This question was asked in our "iit-jee grand test".!

Did they expect us to solve it by this method or is there any other simple method too (which is within jee level) ?

For the inequality what is needed is an inequality known as Hölder Inequality. It is a generalisation of Cauchy Schwarz inequality

I will give it in a form that is useful in this problem:

\left(x_{11}^3 + x_{12}^3 + x_{13}^3 \right)^{\frac{1}{3}} \left(x_{21}^3 + x_{22}^3 + x_{23}^3 \right)^{\frac{1}{3}} \left(x_{31}^3 + x_{32}^3 + x_{33}^3 \right)^{\frac{1}{3}} \ge \left(x_{11}x_{21}x_{31} + x_{12}x_{22} x_{32} + x_{13}x_{23}x_{33} \right)

So when you apply it, we get

\left(8x^2+y^2+z^2 \right) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) \ge (2+1+1)^3 = 64

Note: In JEE Cauchy Schwarz itself may not find any application. So Hölder may be overkill as far as JEE preparation is concerned.

The answer given is 64 itself, but most of my friends argued it to be 54.!

i couldn't have a detailed discussion with them, so posted it here if anyone could solve it in an easier/compatible way.!

brother, u post ur generalization thought, it may be right as the answer is correct :)

but please, post it in a detailed way, as i'm newly exposed to this cauchy schwarz !

@Nishant bro. Thanx for the links. I got the inequality now. :)

but, i didn't get ur soluton yet ! please post 2-3 intermediate steps at the point where u wrote "now apply cauchy schwarz". that will be helpful for everyone, especially me ! please

=50!/6!44! - 5!/4!.40!/6!34! + 5!/2!3!.30!/6!24! - 5!/3!2!.20!/6!14! + 5!/4!.10!/6!4!
=1/6!(50!/44!- 5!/4!.40!/34! +5!/2!3! . 30!/24! - 5!/3!2!.20!/14! + 5!/4!.10!/4!)
=1/6!([50*49*48*47*49*45]-[40*39*38*37*36*34] + [2*5*30*29*28*27*26*25] -[5*2*20*19*18*17*16*15] + [10*9*8*7*6*5])
10/6!([5*49*48*47*49*45]-[4*39*38*37*36*34] +[*30*29*28*27*26*25] -[*20*19*18*17*16*15] + [*9*8*7*6*5]

50/6!([49*48*47*46*45]-[4*39*38*37*7.2*34] +[6*29*28*27*26*25] -[4*19*18*17*16*15] + [*9*8*7*6])
200/6!([49*12*47*46*45]-[39*38*37*7.2*34] +[6*29*7*27*26*25] -[19*18*17*16*15] + [*9*2*7*6])
1800/6!([[49*12*47*46*5]-[13*38*37*2.4*34] +[6*29*7*3*26*25] -[19*6*17*16*5] + [**2*7*6])
=14400/6!([49*3*47*23*5]-[13*38*37*0.3*34] +[3*29*7*3*13*12.5] -[19*6*17*2*5] + [3.5*3])

i am getting mad

u r correct :)

solve the other two also. please!