Q = a 3 ( b + c ) 1 + b 3 ( a + c ) 1 + c 3 ( a + b ) 1
Let a , b , and c be positive numbers such that a b c = 1 .
What is the minimum value of the above expression?
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Very good!!
Excellent solution.
How do you get the 2 ( a b + b c + c a ) ≥ 3 2 3 ( a b c ) 2 ???
Can you please explain how you got third line from second line ? Thanks.
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The third line is nothing but the simplified version of the fraction's numerator.
Edit: Please don't add please while addressing us, Sir. Your wish is our command :)
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I appreciate you sentiments. However I know how more intelligent and knowledgeable you youngsters are. I feel very happy at that.
It is funny how I read the line between numerator and denominator in place of equal to sign when I had put the question to you !! Yes it is perfectly clear.
With best wishes.
I Used aM -Gm .Thats easy
As abc=1 & a,b,c > 0, let a= x/y >0, b= y/z >0, c= z/x >0. Then Q= 1/((x/y)^3(y/z + z/x)) + 1/((y/z)^3(z/x + x/y)) +1/((z/x)^3(x/y + y/z))
= 3[(x^3y^3z^3/x^3y^3z^3(y/z + z/x)(z/x + x/y)(x/y +y/z)]^(1/3) =3[ (y/z + z/x)(z/x + x/y)(x/y +y/x)]^(1/3) as AM >= GM and the equality occurs iff y/z + z/x = z/x + x/y = x/y + y/z iff y/z = z/x = x/y (each=[yzx/zxy]^(1/3) =1) iff x= y =z = 1 Hence Q >= 3[1/2 1/2 1/2]^(1/3)=3/2. So the minimum value of Q is 3/2. Note: when I read the problem,first idea strikes a=b=c=1 & min.Q=3/2. Then I followed as above but forgot to find the cube root of 8 & got min.Q=3/8. I wonder but did not check & selected the answer 3/8 =.375 with full doubt!...
Another way using AM-GM. Call the expression A, we have a 3 ( b + c ) 1 + 4 a ( b + c ) ≥ a 1 = b c Doing the same with the other two then combine all three inequalities we get A + 2 a b + b c + a c ≥ a b + b c + a c ⇔ A ≥ 2 a b + b c + a c ≥ 2 3 3 ( a b c ) 2 = 2 3 = 1 . 5 The equality holds when all the variables equal to 1
A very simple way is to apply AM-GM inequality for 3 reals a,b and c. We can get as a=b=c=1
another simple way...
to get a minimum value the three numbers should be natural numbers rather than decimal numbers
since abc =1
therefore a=b=c=1
then q= 3/2 =1.5
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We are given the condition positive numbers.Not natural numbers. Value 1 holds since abc equals it.. and u r required to find minimum..But this does not holds in most conditions.. Try this one https://brilliant.org/problems/1-more-day-to-valentine/?group=HKTdv8hnP4Rh
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all natural numbers are positive numbers
according to this question , I am insisting that a natural number can only give a minimum value...
you are right that it does not hold in most conditions;
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@Aswin T.S. – You will then need to substantiate why in this particular case, "only a natural number can give a minimum value". It is not immediately obvious why this is true.
I seem IMO inequality in 1995 so easy =)))
Let a=1, b=1, and c=1. Clearly, Q=1.5. Now code -- in your language of choice -- a loop as follows:
for (int i=0; i<100; i++) { a = .9999999a; b = a; c = 1.0/(ab); Q = 1.0/(a^3(b+c)) + 1.0/(b^3(a+c)) + 1.0/(c^3(a+b)); }
Q grows larger as 'a' grows smaller. Since a, b, and c are dependent on each other (abc = 1), we'll get the same result for b or c getting smaller. Q always grows larger.
Therefore, Q=1.5 is the minimum.
One can invoke the minimum-maximum inequality, and let a=b=c. Now, we are given the constraint equation, abc = 1, and by invoking min-max, we yield that a^3 = 1 which means that a must equal 1. Now, by setting a=b=c, we are trying to determine whether we yield a max or min. By plugging in a trial solution on say, a=1/3, b=3, c=1, we see that the result is large compared to when we set a=b=c=1. Therefore, a=1 must be a minimum and by plugging into the expression given, we yield 3/2 as desired.
One can also utilize Lagrange Multipliers to bash this expression as it is in terms of f(a,b,c). I will leave that up to the reader.
It is a symmetric expression. So likely solution is a=b=c=1. Then Q=1.5. Let us investigate what happens if it is not so.
Since a
b
c=1, we change a, and b keep c=1. Let us say a= 1+x for a positive x.
∴
b
=
1
+
x
1
.
Q
=
(
x
+
1
)
3
∗
(
x
+
1
1
+
1
)
1
+
(
x
+
1
)
3
1
∗
(
x
+
1
+
1
)
1
+
1
∗
(
x
+
1
1
+
(
1
+
x
)
)
1
Q
=
(
x
+
1
)
2
∗
(
x
+
2
)
1
+
x
+
2
(
x
+
1
)
3
+
x
2
+
2
x
+
2
x
+
1
There are two possibilities. (a) x
≥
1, (b) x< 1.
(a) If x
≥
1, middle term alone ...........Q>2>1.5.
(b) If x< 1, let is evaluate Q again. What ever x we take as x decreases from 1 to zero Q also decreases,
but remains > 1.5. It is not a nice proof. Any one can please give a better ?
In same way we can prove when all there are not 1, though the proof would be messy.
You have not considered the cases where none of a , b , c is equal to 1.
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Yes. As I have completed my solution above, the proof becomes messy so did not continue. I think there can be a better way some one can give.
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One possible way to complete your solution along the approach that you took, is to show that if a ≥ 1 ≥ b , then
f ( a , b , c ) ≥ f ( 1 , a b , c ) ≥ f ( 1 , 1 , 1 ) .
Note: I haven't checked if this would work.
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The very expression can be rewritten as a b + a c a 2 1 + b a + b c b 2 1 + c a + c b c 2 1 .
By Titu's Lemma, Q ≥ 2 ( a b + b c + c a ) ( a 1 + b 1 + c 1 ) 2 .
Since a b c = 1 , ( a 1 + b 1 + c 1 ) 2 = ( a b + b c + c a ) 2 .
Hence, the given expression is ≥ 2 ( a b + b c + c a ) ≥ 3 2 3 ( a b c ) 2 = 2 3 .