11SIR FIR AMIT KE RASTE PE CHALTE HAI
BUT MATHEMATICALLY I DONT HAVE ANYTHING IN MY MIND ELSE
I WILL BE GLAD TO SEE UR POINT OF VIEW[1]
I wiill put up some good questions of AM GM inequality only...
So that you all can try them.. some of themwill be very very easy.. others will be a bit more tough...
So try them all one by one..
I will try to post them in order of difficulty.
Easy one liners:
A1) \frac{a+b}{a}+\frac{b+c}{b}+\frac{c+a}{c}
A2) \frac{a-b}{a}+\frac{b-c}{b}+\frac{c-a}{c}
A3) Prove: a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)>6abc
A4) Prove : AM HM inequality for 3 numbers a,b,c
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68 Answers
11@PROPHET SIR
I DON'T UNDERSTAND THE CONCEPT OF ASKING THIS QUES[2]
BUT I KNOW THAT
D=-B+-√B2-4AC/2A
SO
its roots r 1 and 2
we can also see this from the graph
but sir what new have to prove here????
341Now the plot gets thicker (honeymoon's over guys!):
If a,b,c>0 and \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \ge 2, prove that abc \le \frac{1}{8}
11OOOOOHHHHHHHHHHHHHHHH
SIR AAP NE TO DIL JEET LIYA[8]
SIR THIS IS SIMPLE AND THE BEST SOLUTION EVER SEEN
SIR BAHUT BADDIYA
THANKS 4 SHARING[1]
341Chalo theek hey
1+a \ge 2 \sqrt a etc.
Hence (1+a)(1+b)(1+c) =8 \ge 8 \sqrt {abc} and the conclusion follows.
Amazingly in most of the textbooks you will find the same approach as given by manipal.
11341haan lekin usse bhi chota (choti? my hindi is rudimentary!} solution nikal sakte hein
11sir ye waala use AM>GM
a+b+c/3>3√abc
ab+bc+ac/3>3√a2b2c2
now ut abc =t
u will get the answer [1]
(sir pehle kia hua hai!!!!!!!!!!!!)
1taking a=b=c=1 then only the equation becomes 1 hence abc=1
which the maximum value, thus for values of a,b,c for which
(1+a)(1+b)(1+c) is less than 8 the product of abc will be less than 1
341arre bhai, there is a post (No.34) after that clarifying what I wanted to say. I got muddled over this one somehow
Ok, now one more:
If (1+a)(1+b)(1+c) = 8, prove that abc ≤ 1
11(honeymoon's over guys!): [3][3][3] sir abhi to shaadi hi nahin hui[2][2][2]
sir main to #47 ki tarah hi proceed karke hi abswer nikaal leta
but u r a genius
so lt me think ur ways coz i bet u r having a shorter solution
1Nishant . about the one for general thing or C4 ??
method correct statement what does it refer.
and for the general on efurther kya hota hai . bhool gaya hoon plz continune
1for the general one. for n numbers.
let the AM of the n numnbers be A .
we need to add A some specific number of times and then proceed . Nishant Bhaiyya . please batao . naa . please ...
62I think prophet had given this for the same reason that people generally dont see simple things :)
1easy done feeling good to have perhaps solved prophets sirs questions with such ease
im sure he gave it only for time pass
11sir fir meri bhi to post # 74 theek hai na
chalo lets think on ur inequality again[12][12]
1Reviving old threads -
C . 4 > 1 + a1 + a123 ≥ a1 ....................By plain AM - GM
Similarly we have to write down " n " expressions and multiply them out .
We get the required inequality .
C . 5 > 1 + 2 + 2 2 + 2 3 + .............2 n - 1n ≥ ( 1 . 2 . ...........2 n - 1 ) 1 / n = 2 n ( n - 1 ) / 2 n = 2 n - 1
So , 2 n - 1n ≥ 2 n - 1
3413(a^4+b^4+c^4) \ge (a^3+b^3+c^3)(a+b+c) \ge 3abc(a+b+c)
The first inequality is Chebyshev Inequality and the second one is of course AM-GM
11App 1)
P(x) = x2 - 3x +2
= x2-2x-x+2
=x(x-2)-1(x-2)
= (x-1)(x-2)
For
(x-1)(x-2)>=0
x >=1
x>=2
Since x>0 therefor p(x)>=0
Pls check sir
1show that
a+b+c\leq \frac{a^{3}}{bc}+\frac{b^{3}}{ac}+\frac{c^{3}}{ab}
and find when equality can hold.
341\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge 2 \Rightarrow \frac{1}{1+a} \ge \left(1 - \frac{1}{1+b} \right) + \left(1 - \frac{1}{1+c} \right)\\ \\ = \frac{b}{1+b} + \frac{c}{1+c} \ge 2 \sqrt{\frac{bc}{(1+b)(1+c)} by AM-GM
Thus we have the three inequalities
\frac{1}{1+a} \ge 2 \sqrt{\frac{bc}{(1+b)(1+c)}} \\ \\\frac{1}{1+b} \ge 2 \sqrt{\frac{ca}{(1+c)(1+a)}} \\ \\ \frac{1}{1+c} \ge 2 \sqrt{\frac{ab}{(1+a)(1+b)}}
Multiplying and cancelling common elements on both sides, we get
abc \le \frac{1}{8}
1a^3 + 1 +1 mein lagado aajayega
a^3 + 1 +1 > = 3a
ho gaya [4][4]
341@dimensions-yup thats right. but there is an easier way. Which means the others need not stop trying
1@ rahul
given
(1+a)-1+(1+b)-1+(1+c)-1≥2
& by ur AM HM inequility
(1+a)-1+(1+b)-1+(1+c)-1≥(9/(a+b+c+3))
so if x≥y
& x≥z
then we cannot conclude that y≥z or z≥y !!!!!!
1let a,b,c be roots of the equation
x3-s1x2+s2x-p=0
here
s1=a+b+c
s2=ab+bc+ac
p=abc
so the cubic with the roots (1+a)-1,(1+b)-1,(1+c)-1will be
(1+s1+s2+p)x3-(3+2s1+s2)x2+(s1+3x)-1=0
so we r given that the sum of roots of the above equation is greater than 2
i.e. (3+2s1+s2)≥2(1+s1+s2+p)
ab+bc+ac≤(1-2abc).....(i)
by AM GM we have (ab+bc+ac)≥3(abc)2/3.....(ii)
so we have (1-2abc)≥3(abc2/3)
8(abc)3+15(abc)2+6(abc)-1≤0
(abc+1)3(8abc-1)≤0
so we have
abc≤(1/8)
is this method correct BHATT SIR...