24wrong dbt raised...sorry for that
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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37 Answers
1q)9
squaring both sides of inequality
we get
(a.2x+b.3y+1)2≤(4x+9y+1)(a2+b2+1)
a2.22x+ b2.32y + 1 + 2ab.2x.3y + 2b3y + 2a.2x
≤ a2.4x + b24x + 4x + a29y + b29y + 9y + a2 + b2 + 1
implies
a2.4x + b29y + 1 + 2ab.2x.3y + 2b.3y + 2a.2x
≤ a2.4x + b24x + 4x + a29y + b29y + 9y + a2 + b2 + 1
implies
(a29y-2ab.2x.3y + b24x) +(4x - 2a.2x + a2 ) + (9y - 2b3y + b2) ≥ 0
(a3y-2xb)2 + (2x-a)2 + (3y-b)2 ≥0
hence the original inequality is true
1no eureka q cant be <1 because then
1/p+1/q will be more then 1
and not equal to 1
46 ) From the 3 inequalities a2 > bc ( 1 )
b2 > ac ( 2 )
c2 > ab ( 3 )
Dividing 1/2 , 2/3 , 3/1 we get a > b >c >a Not possible!!!!!
24@arshad..
one mistake in Q10
u worte 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)
It should be
(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 times) + (br/s * br/s *......+xs times))1/(ry+xs)
so that ((ax/y + ax/y+..........ry times) + (br/s + br/s +.......+xs times))/(ry+xs)≥((axr)*(bxr))1/xr
=> (.....) ≥ab
48 ) Considering 3( ab + bc + ca) ≤ (a+b+c)2
<=> (a2 + b2 + c2 ) ≥ ab + bc + ca
which is always true ! (By a standard Theorem)
Considering (a+b+c)2 ≤ 4(ab+bc+ca)
<=> a2 + b2 + c2 ≤ 2(ab + bc + ca)
By using Cosine formula
(a2 + b2 + c2 ) = 2abcosC + 2bccosA + 2cacosB
≤ 2( ab + bc + ca)
Hence a2 + b2 + c2 ≤ 2(ab+bc+ca)