71Rest can also be done. I don't think they need anything more than AM-GM. Just try to rearrange them.
1) If a,b,c,d > 1
P.T. 8(abcd+1) > (a+1)(b+1)(c+1)(d+1)
2) If a,b,c ≥ 0 and a+b+c =1
Find the max. value of \sqrt[ 3 ]{a+b} + \sqrt[ 3 ]{b+c} + \sqrt[ 3 ]{a+c}
3) For a,b,c > 0 Prove that
x2 + y2x+y + z2 + y2z+y + x2 + z2x+z ≥ x+y+z
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14 Answers
71For second, Is the max. value 3 ?
I did it like this
\sqrt[ 3 ]{a+b} => 3√1-c
\sqrt[ 3 ]{b+c} => 3√1-a
\sqrt[ 3 ]{a+c} => 3√1-b
Clearly, It'll be max when a.b.c are min.
i.e., a=b=c=0.
Therefore Max. value = 1+1+1 = 3.
71
1It is given that a+b+c=1.
If a=b=c=0 how is this possible? [3]
And 3√a+b ≤ 3√1-c if they are 0.....
Could you please show the working for the 1st sum
21for the 2nd one...dunnoh its ri8 or wrong..
let f(x)= x1/3
its double derivative is negative for x>0
now from jensens inequality {(a+b)1/3 + (b+c)1/3+(a+c)1/3}/3 ≤ {2(a+b+c)/3}1/3 (its also that mth power of AM≤AM of mth power)
given a+b+c =1 ,so max value is 3*(2/3)1/3
this also follows the fact that for max value terms has to be equal ie a=b=c=1/3
21yaar i confused my self...
it can also be done by AM of m th power≥m th power of AM
213rd one is very simple...
from AM≥GM
(x2+y2 )/(x+y)≥(x+y)/2
now it can be done///
21sir,not getting the proof of the hint...it seems obvious but cant prove it...
341(x-1)(y-1)>0 and hence xy+1>x+y and hence
2xy+2>xy+x+y+1 = (1+x)(1+y)
So 2(xy+1)>(1+x)(1+y)
12(xy+1)>(1+x)(1+y) as sir said.
then taking x=ab,y=cd [as a,b,c,d>1,ab>1,cd>1]
2(abcd+1)>(1+ab)(1+cd) ..(1)
also 2(ab+1)>(a+1)(b+1)
and 2(cd+1)>(c+1)(d+1)
multiply (1) by 4 on both sides:
8(abcd+1)>[2(ab+1)][2(cd+1)]
or 8(abcd+1)>(a+1)(b+1)(c+1)(d+1)
1Thanx a lot prophet sir and arka....
Could u plzzzzz help me on the 2nd?
