inequalities starter-1(plz give as simple a solution as possible)

By 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.

By 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

(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)

thank you!

the trig formula would be applicable for the expression b-a/1+ab

yeah...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 ?)

proof: 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.

i 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

actually 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]

@shubhodip
i know how it feels[3]
i die a death daily @wikipaedia[2]