TargetIIT

If ax[p]2[/p] - bx + c = 0 have two distinct roots lying in the interval (0,1), a, b, c ε N, then prove that log[ss]5[/ss](abc) ≥ 2.

sum of the roots = b/a
prod. of roots = c/a
AM>=GM>=HM

SO
4ac>=2c/b
a+c/2>=4ac

this is just the funda i cud reach
help me out!!!!

I dont think that will be of much help....
Hint: f(x)=a(x-α)(x-β) where α,β ε (0,1)
so, f(0)f(1)=a[p]2[/p]αβ(1-α)(1-β)
AM≥GM will be helpful...now proceed...

then i suppose
f(0)f(1)<0 so that it may hv 2 real roots btwn (0,1)

totally wrong....read the question carefully...
both the roots here lie between 0 and 1. So, f(0)f(1) > 0 always.

srry!! i mistook the funda --- the pt. at which the tangent =0 & then looking for sign change..for the orig. function....

no one seems to have solved this one..!!!

I'm trying hard but not getting it......any small hint ?

I'm getting c > 0

and

a-b+c > 0

read the question again...it is already given that a,b,c are natural numbers. Try it a little longer , i think u will get it...
0<α<1
so,  0<(1-α)<1
now proceed...use AM≥GM

See this only if you are desperate: [hide]  http://goiit.com/posts/list/algebra-problem-no-3-67916.htm#336240[/hide]

But there is a much easier way i saw later. So prob is still open

Sir, what is the easier method?

(getting desperate to see that[1])

Since the polynomial [im]http://codecogs.izyba.com/gif.latex?f%28x%29%3Dax%5E2-bx+c[/im] has real and distinct roots,
b[p]2[/p] > 4ac
Also, [im]http://codecogs.izyba.com/gif.latex?f%28x%29%3Da%28x-%5Calpha%29%28x-%5Cbeta%29[/im]
Since the roots of f lie strictly between 0 and 1 so
[im]http://codecogs.izyba.com/gif.latex?f%280%29f%281%29%3E0[/im]
But being an integer, f(0)f(1) ≥1
Also
f(0)f(1)=
[im]http://codecogs.izyba.com/gif.latex?a%5E2%5Calpha%281-%5Calpha%29%5Cbeta%281-%5Cbeta%29%3C%5Cdfrac%7Ba%5E2%7D%7B16%7D[/im]
In this last step, I used AM-GM and the fact that equality cannot hold.
Combining the inequalities we get
[im]http://codecogs.izyba.com/gif.latex?a%5E2%3E16%5CRightarrow%20a%5Cgeq%205[/im]
Thus,
[im]http://codecogs.izyba.com/gif.latex?b%5E2%3E20c%5Cgeq%2020[/im]
which gives us
[im]http://codecogs.izyba.com/gif.latex?b%5Cgeq%205[/im]
I have taken c≥1.
Hence,
[im]http://codecogs.izyba.com/gif.latex?abc%5Cgeq%2025[/im]
With a=5, b=5, c=1, the conditions of the problem are indeed satisfied, so abc has a minimum value 25.

Thanks sir.