IMO 1995

Algebra Level 2

Q = 1 a 3 ( b + c ) + 1 b 3 ( a + c ) + 1 c 3 ( a + b ) Q=\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}

Let a , b , a,b, and c c be positive numbers such that a b c = 1 abc = 1 .

What is the minimum value of the above expression?

0.25 5 3 1 2 1.75 0.375 1.5

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7 solutions

Swapnil Das
Feb 15, 2016

The very expression can be rewritten as 1 a 2 a b + a c + 1 b 2 b a + b c + 1 c 2 c a + c b \dfrac{\frac{1}{a^{2}}}{ab+ac}+\dfrac{\frac{1}{b^{2}}}{ba+bc}+\dfrac{\frac{1}{c^{2}}}{ca+cb} .

By Titu's Lemma, Q Q ( 1 a + 1 b + 1 c ) 2 2 ( a b + b c + c a ) ≥\dfrac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}}{2(ab+bc+ca)} .

Since a b c = 1 abc=1 , ( 1 a + 1 b + 1 c ) 2 = ( a b + b c + c a ) 2 (\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^{2}=(ab+bc+ca)^{2} .

Hence, the given expression is ( a b + b c + c a ) 2 3 ( a b c ) 2 3 2 = 3 2 ≥\dfrac{(ab+bc+ca)}{2}≥3\dfrac{\sqrt[3]{(abc)^{2}}}{2}=\dfrac{3}{2} .

Very good!!

Adarsh Kumar - 5 years, 4 months ago

Excellent solution.

ZK LIn - 5 years, 4 months ago

How do you get the ( a b + b c + c a ) 2 3 ( a b c ) 2 3 2 \dfrac{(ab+bc+ca)}{2} \geq 3\dfrac{\sqrt[3]{(abc)^{2}}}{2} ???

AccelNano Lim Loong - 5 years, 3 months ago

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Am-Gm on (ab+bc+ca)

Department 8 - 5 years, 3 months ago

Can you please explain how you got third line from second line ? Thanks.

Niranjan Khanderia - 5 years, 4 months ago

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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 :)

Swapnil Das - 5 years, 4 months ago

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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.

Niranjan Khanderia - 5 years, 4 months ago

I Used aM -Gm .Thats easy

Kaustubh Miglani - 5 years, 4 months ago

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!...

vinod trivedi - 5 years, 3 months ago
P C
Feb 15, 2016

Another way using AM-GM. Call the expression A, we have 1 a 3 ( b + c ) + a ( b + c ) 4 1 a = b c \frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq\frac{1}{a}=bc Doing the same with the other two then combine all three inequalities we get A + a b + b c + a c 2 a b + b c + a c A+\frac{ab+bc+ac}{2}\geq ab+bc+ac A a b + b c + a c 2 3 ( a b c ) 2 3 2 = 3 2 = 1.5 \Leftrightarrow A\geq\frac{ab+bc+ac}{2}\geq\frac{3\sqrt[3]{(abc)^2}}{2}=\frac{3}{2}=1.5 The equality holds when all the variables equal to 1

Sridhar Sri
Feb 16, 2016

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

Aswin T.S. - 5 years, 3 months ago

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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

Sridhar Sri - 5 years, 3 months ago

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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;

Aswin T.S. - 5 years, 3 months ago

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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.

Calvin Lin Staff - 5 years, 3 months ago
Son Nguyen
May 10, 2016

I seem IMO inequality in 1995 so easy =)))

Richard Levine
Mar 19, 2016

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 1 + x . Q = 1 ( x + 1 ) 3 ( 1 x + 1 + 1 ) + 1 1 ( x + 1 ) 3 ( x + 1 + 1 ) + 1 1 ( 1 x + 1 + ( 1 + x ) ) Q = 1 ( x + 1 ) 2 ( x + 2 ) + ( x + 1 ) 3 x + 2 + x + 1 x 2 + 2 x + 2 \therefore ~b=\dfrac 1 {1+x}.\\ Q=\large \dfrac 1 {(x+1)^3*(\frac 1 {x+1} +1)} +\dfrac 1 {\frac 1 {(x+1)^3}*(x +1 +1)} +\dfrac 1 {1*(\frac 1 {x+1}+(1+x))}\\ Q= \large \dfrac 1 {(x+1)^2*(x +2)}+\dfrac {(x+1)^3}{x +2} +\dfrac {x + 1} {x^2+2x+2}
There are two possibilities. (a) x \geq 1, (b) x< 1.
(a) If x \geq 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 a, b, c is equal to 1.

Calvin Lin Staff - 5 years, 3 months ago

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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.

Niranjan Khanderia - 5 years, 3 months ago

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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 a \geq 1 \geq b , then

f ( a , b , c ) f ( 1 , a b , c ) f ( 1 , 1 , 1 ) . f( a, b, c) \geq f ( 1, ab, c ) \geq f (1, 1, 1).

Note: I haven't checked if this would work.

Calvin Lin Staff - 5 years, 3 months ago

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@Calvin Lin Thank you. I will try.

Niranjan Khanderia - 5 years, 3 months ago

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