Some doubts.!

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

hmmm....gud yaar...i think its now clear !!!

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 !?

I think he meant at what values do we get the value 54?

the values of x,y,z

surely the inequality u've used is far more sound mathematically...but whr is the concept going wrng with AM - GM thing ??

see...

(8x² + y² + z²)/3 ≥ (8x2 * y2 * z2 )^1/3
now calculating..v get :

(8x2 + y2 + z2) ≥ 6 (xyz)^2/3 ... (1)

Second part :

(x^-1 + y^-1 + z^-1 )/3 ≥ (1/xyz)^1/3

again shfting 3 to rite and squaring...v get

(x-1 + y-1 + z-1 )² ≥ 9 (1/xyz)^2/3 ... (2)

multiplying (1) and (2) we get the result naa...strangely it comes out to be independent of x, y and z...
plzz sum1 temme wats wrong...

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.

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