4Plzzzzz correct me if I'm wrong!!!!!!!!
It all started with b555's classic integration siikho thread..and now nishant sir started some more threads on it....
I ma starting one small on inequalities..A lot of inequlaities thread are already there on site..I know that...but this is just my contribution [1][1]
I guarantee all these are JEE level only and no olympiad level will be posted as far as possible [6]
ALso i odnt have copyright for ""sikho"" word..so if b555 wants he can sue me [3][3]
some of these questions will be really easy..but htas how JEE inequalitites willl be [6][6]
italiciised questions mean answered questions.....[1][1]take care of this[1]
Q1 Prove 12√n+1<12.34.56.......2n-12n< 1√2n+1
Q2 If a,b,c>0 and b+c>a, prove that a1+a<b1+c+c1+b
Q3 For real nos. a,b,c,d prove that (1+ab)2+(1+cd)2+a2c2+b2d2≥1
Q4 Prove that 3b2+6ac≤(a+b+c)2,if a≥b≥c
Q5 Non negative nos a,b,c,d such that a+b+c+d=1,prove that ab+bc+cd≤1/4
Q6 Prove that inequalities involving following finite quantities a2-bc(a+b)(a+c)>0,b2-ac(a+b)(b+c)>0,c2-ab(a+c)(b+c)>0
cannot be simultaneoulsy true
Q7If a,b,c, are real positive quantities,show that
1a+1b +1c ≤ a8+b8+c8a3b3c3
Q8 If a,b,c are sides of triangle then prove that 3(ab+bc+ca)≤(a+b+c)2≤4(bc+ca+ab)
Q9 If a>0 ,b>0 the nfor any real x and y ,prove taht following inequality holds true
a.2x+b.3y+1≤√4x+9y+1√a2+b2+1
Q10 If a,b,c are positive nos. and p,q be rational nos.(p>1) such that 1/p +1/q=1,tehn prove that ab≤app+bqq
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UP 0 DOWN 0 4 37
37 Answers
3q2)a<b+c
a/1+a<\frac{b+c}{b+c+1}
a/1+a<\frac{b}{b+c+1}+\frac{c}{c+b+1}
a/1+a<\frac{b}{b++1}+\frac{c}{c++1}
4Theorem i mentioned in Q8
If n>1 , ai > 0 for 1≤ i ≤ n & S = a1 + a2 + ..an ,then
sigma (i=1 to i=n) 1/(s-ai) ≥ n2/(n-1)s
& sigma ai / (s-ai) ≥ n/(n-1)
1q)8
in triangle
a+b>c (common fact)
b+c>a
and
c+a>b
am≥gm
(b2 + c2)/2≥ bc........1
(a2 + c2)/2≥ ac......2
(b2 + a2)/2≥ ba............3
1+2+3
a2 + b2 + c2 ≥ab+bc+ca
adding 2(ab+bc+ca) on both sides
(a+b+c)2≥3(ab+bc+ca)..........4
in triangle ABC
a2 = b2 + c2 - 2bc cosA
b2 + c2 - a2 =2bc cosA≤2bc
b2 + c2 - a2≤2bc....5
c2 + a2 - b2≤2ca.....6
a2 + b2 - c2≤2ba.............7
5+6+7
a2 + b2 + c2≤2bc+2ac+2ab
adding 2bc+2ac+2ab
(a+b+c)2≤4(ab+bc+ca).....8
so from 4 and 8
so hence proved..........
24For everyone's convenience....I have italiciised questions answered questions.....[1][1]
so that no one wastes his time typing the already answered question here.....u may solve it on ur own in ur notebook though[1]
24are all the ques solved???????????????????
i dont think so...then why have the discussions stopped????
19i am going to try the left out ones.....eure..please mention the ones left out !
24ya sure it was mistyping..becoz u got the rite ans....[1]
@uttara...
u assumed
LHS ≥ 1
how can u assume something u have to prove??
14.
we need to prove 3b^2+6ac \le a^2+b^2+c^2+2ab+2bc+2ca
or 2b^2+4ac \le a^2+c^2+2ab+2bc
from AM-GM we have a^2+c^2 \ge 2ac
so we need to show
b^2+2ac \le ab+bc+ac
or b^2+ac \le ab+bc
or b(b-a) \le c(b-a)
or c\le b (since b-a \le 0)
which is given
1is'nt 3 very obvious
(1+ab)2 ≥1 all other terms on the LHS are squares so they are also positive
so their sum is also positive hence LHS ≥1
1itz nt as obvious coz a,b belongs to real no.'s
so jst as an example let a=1 and b=-1
then ur consideration doesn't hold
24Dont use calculus dude,,,,,thes can be solved easily with algebra..[1]
43 ) LHS = 2 + a2b2 + c2d2 +a2c2 + b2d2 + 2 (ab + cd)
= 2+ ( a2+d2) (b2+c2) + 2(ab+cd)
≥ 1 + (ab + cd)2 + 2(ab +cd) [(ab +cd)2 ≤ (a2 + d2)(b2 + c2 ) Using Cauchy Inequality ]
≥ (ab + cd + 1)2
≥ 0
which is always true
Hence proved
1let me start(wanted some junior to prove but no ones coming forward)
q1.
let s=1/2*3/4*5/6.........(2n-1)/2n<2/3*4/5*6/7.......2n/2n+1
(as p/q<p+1/q+1)
s<2/3.4/5.......2n/2n+1
now shifting denominator 1 place we get
s<1/s(2n+1)
s^2<1/2n+1
s<1/√(2n+1)
rhs proved
also
(2n+1)s=(2n+1)(1/2.3/4.5/6........2n-1/2n
=3/2.5/4.7/6.....(2n-1)/2n>4/3.6/5.8/7.....2n+2/2n+1
therefore
(2n+1)s>n+1/s(2n+1)
(2n+1)^2(s^2)>n+1
s^2>n+1/(2n+1)^2
so
s>√(n+1)/2n+1
so
s>1/2√(n+1)
(as √n+1/2n+1>1/2√(n+1)
so lhs also proved
hence q1 solved
1q)7
(a+b+c)/3>(abc)1/3
(a+b+c/3)6>a2b2c2..........1
also
a2+b2/2>ab......2
and
b2+c2/2>bc......3
and also
a2+c2/2>ac...4
now 2+3+4
a2+b2+c2>ab+bc+ca
(a+b+c)2>3(ab+bc+ca)
(a+b+c/3)2>1/3(ab+bc+ca).......5
and also
a8+b8+c8/3>(a+b+c/3)8.......6
1*5
(a+b+c/3)8>a2b2c2(ab+bc+ca)/3.....7
and
6*7
a8+b8+c8>a3b3c3(1/a+1/b+1/c)
therefore
a8+b8+c8/a3b3c3>(1/a+1/b+1/c)
hence proved.........
i knw that eure would want others to do such questions but still no one has replied from a long time....so i did
1Q4
(a+b+c)2=b2 +2ac+ (a2+c2)+2b(a+c)
a2+c2>=2ac
a<b<c
a+c>b
using inequality
(a+b+c)2>=3b2+4ac
wer am i wrong
62@xyz.. you are not wrong.. only that you have not proved what has to be...
@Arshad.. good work... btw just prove your statement 6 in #8 *It is correct.. but just explain why you said that?
1@ nishant sir
i said it because arithmetic mean of nth powers>nth power of arithmetic mean
1now q 10)
p>1
and 1/p + 1/q=1
implies
q>1
let p=x/y
and q=r/s
a,b,x,y belong to I+
now
ap/p + bq/q=(ax/y)/(x/y) + ( br/s)/(r/s)
=(ry ax/y + xs br/s)/xr...........1
(ry ax/y + xs br/s)/xr=((ax/y + ax/y+..........ry times) + (br/s + br/s +.......+xs times))/xr......2
but
1/p + 1/q=1
ry+xs=xr.............3
frm 2 and 3
(ry ax/y + xs br/s)/xr=((ax/y + ax/y+..........ry times) + (br/s + br/s +.......+xs times))/(ry+xs)
≥((ax/y * ax/y*..........*ry factor) + (br/s * br/s *.......*xs factor))1/(ry+xs)
≥ (axr bxr)1/xr
≥ ab............4
and from 1 and 4
we get
ap/p + bq/q≥ab
hence prooved