1A.M. >= G.M.
so, (x+y+z)/3 >= (xyz)^1/3
=> xyz = (6/3)^3 =2^3=8.
now,,, 1/3 * (a/x + b/y + c/z) >= (abc/xyz)^1/3
=> 1/3 * (a/x + b/y + c/z) >= (abc)1/3/2
=> (a/x + b/y + c/z) >= 3(abc)1/3/2
hence min value .... 3(abc)1/3/2.
Easy if u have seen. Try this even if u havent. It is a good one to know :)
x+y+z=6
all are positive reals!
Find the minima and maxima of
a2/x + b2/y + c2/z
where a, b, c are integers!
Really Sorry I made a mistake in this question (Sky and Sharat.. )
This was the expression i gave initially.. a/x + b/y + /z
The explanation is there for this question.. but not a proper proof with me.. may be someone would want to take a cue from the proof of this question....
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15 Answers
1
62hmm..
good one.. :)
neone else with some other method.. that may be just a tad smaller?
1x + y + z = 6
xyz will be maximum when x=y=z.
Therefore , xyz is max when x=y=z=2.
Now the given exp = (yza + xzb + xyc)/xyz
It will be min when xyz is max.
therefore substitue x=y=z=2.
The exp becomes - (a+b+c)/2
So that is the minimum value of a/x + b/y + c/z
62so ur answer is (a+b+c)/2??
no dear! unfortunately not..
uve made a mistake in assuming "It will be min when xyz is max. "
1I meant the given exp becomes a+b+c/2 and can be used in A.M > = G.M and get the above answer .. but I think most of those steps were unnecessary..
Until xyz = max at x = y = z = 2 would have been enough i think.
62See the question above..
"Really Sorry I made a mistake in this question (Sky and Sharat.. )
This was the expression i gave initially.. a/x + b/y + /z
The explanation is there for this question.. but not a proper proof with me.. may be someone would want to take a cue from the proof of this question...."
btw ur answer is not correct still!
1A.M. >= G.M.
so, (x+y+z)/3 >= (xyz)^1/3
=> xyz = (6/3)^3 =2^3=8.
now,,, 1/3 * (a2/x + b2/y + c2/z) >= [(abc)2/(xyz)]^1/3
=> 1/3 * (a2/x + b2/y + c2/z) >= (abc)2/3/2
=> (a2/x + b2/y + c2/z) >= 3(abc)2/3/2
hence min value .... 3(abc)2/3/2.....
isnt this right??? ...coz it doesnot matter if we are squaring the integers .... i will still be an integer only...
(correct me if wrong...)
1no no... i didn thot only... i jus' asked u whether the previous one wid few modifications will be right or not.... i didn think abt this prob till now (sorry) but i iwll do it widn today.. definitely...
62Hey I had thought that this is not the toughest of the problems.. I expected a few ppl to have done this one!
:(
Still keep trying...
62Bad bad.. i thought someone whould have come up with a solution to this one!! I had put tougher problems which you guys solved!
x/a+x/a+... a times + y/b+y/b+... b times.+ z/c+z/c+.... c times =6
AM= 6/(a+b+c)
HM =(a+b+c)/{ a/x+a/x+... a times + b/y+b/y+... b times.+ c/z+c/z+.... c times}
HM=(a+b+c)/{a2/x + b2/y + c2/z}
Apply AM-HM inequality to these terms,...
AM>=HM
6/(a+b+c)>=(a+b+c)/{a2/x + b2/y + c2/z}
6{a2/x + b2/y + c2/z}/(a+b+c)2>=1
{a2/x + b2/y + c2/z}>= (a+b+c)2/6
Done :)
This is a good trick to have in your sleeve :)
1:(
had thot but didn get so... :(
obviously hav not thot the way u solved....
