problem-1:
a, b be real numbers such that 0 ≤ a ≤ b ≤ 1.
prove:
0 ≤(b − a)/(1 − ab) ≤ 1,
21By Algebra:
0≤b-a1-ab
because we have 0≤b-a, and 0≤1-ab
next we need to prove
b-a1-ab≤1
1-ab is nonnegative, so
b-a≤1-ab
iff,b-1 + ab-a ≤0
iff,b-1 + a(b-1)≤0
iff,(b-1)(a+1)≤0
which is true because of the fact that b≤1
21By trigonometric replacements:
there exists some 450≥θ1≥θ2≥0 , so that tanθ1= b and tanθ2= a
so b-a1-ab = tan(θ1- tanθ2)
It's obvious that 0≤(θ1- θ2)≤450
so 0≤ tan(θ1- tanθ2)≤1
and the inequality follows.
21By calculus:
consider the expression b-a1-ab , Let us choose some 'a'. Now that's a function in 'b'.which is convex. (may be concave also, i have not really checked, but things hopefully don't change )
So maximum value of the expression must occur at endpoints of its domain,i.e. 0 or 1.
Now, let us fix 'b'. This is again convex/concave in 'a'. so maximum/minimum will occur at the end points. i.e. 0 or 1.
hence the expression will take it's maximum/minimum value at (b,a) = (1,1) , (1,0) , (0,1) , (0,0)
Now, (1,1) and (0,1) is abstract because they give 00 form and violate b≥a respectively.
We are left with (b,a) = (1,0) and (0,0)
(b,a) = (1,0) makes the expression equal to 1 and (0,0) makes the expression equal to zero.
so we conclude 0≤ b-a1-ab≤1
21(the abve solution may not be flawless )
This is really a powerful method and solves lot of tough looking problems easily..
btw, download a book 'mathematical olympiad challenges ' by Titu andreescu and Razvan Gelca
(google it, it's easily available)
There is chapter called ''look at the end points''. read that.
If u still hav dbts u may ask here,(not me , the experts :P)
341the trig formula would be applicable for the expression b-a/1+ab
21yeah...was sleeping.;p
also the calculus one is not only wrong but has lot's of mistakes. specially the way i wrote ''maximum/minimum'' ,''things don't change'' is completely wrong. i was nt thinking at all.
actually the thing is , For a convex or linear function maximum value occurs at the end points and for concave functions minimum value occurs at the end points. ( can be proved easily)
@ prophet sir, pls give two links from where i can read and understand karamata's inequality and minkowski inequality.(the minkowski inequality is written with integrations and vectors etc,can it be used in algebra ?)
21proof: for convex function max. value occurs at the end points.
a function is convex in an interval [a,b] ,b>a ,if f ' is increasing.
if f' is non zero , function is either strictly increasing or strictly decreasing on that interval. so maximum value must occur at the endpoints (here the minimum also occurs at the end points)
if f ' is zero at some x = c , then c must be the point of minima.
In the interval [a,c] f ' is negative , so f is strictly decreasing. f(a) > f(x) ,for all x ε (a,c]
in (c,b] it is strictly increasing. so f(b) > f(x) , x ε (c,b]
So the maximum of f is simply max{f(a), f(b)}
in order to check for convex function , f ''(x) >0 for all x in [a,b] is a sufficient condition.
341i have a handout i downloaded from god knows where. If you can send me your email id in my chatbox, i will email it to you. lemme see if i have anything on minkoswski
21actually i have seen somewhere u used minkowski ( most probably in aops) , for minimizing or maximizing some function involving square roots .
but when i searched it in wikipedia , i was killed. [2][17]
