The answer is not 1/9
If a , b , c are non-negative real numbers, what is the maximum value of
( a + b + c ) 3 a b 2 + b c 2 + c a 2 ?
The answer is 0.148148148.
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3 solutions
How do you know to factor it like that without knowing that the max value is 4/27?
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When attempting inequalities, what I first do is try to guess what the maximum value is by plugging in a = b = c , or a = 0 , b = c , or other things like that. Then, when I find a guess, I try to prove it.
Or simply use Mathematica.
That's a good question.
In cases like this, you tend to have to figure out what the answer is first, and look at what the equality case is, before you can proceed to prove it.
Other approaches like Lagrange multiplier, or basic calculus in this case, could give you insight into the answer.
Take a=0 b=1 c=2 and then some it by using the formula given.... U will get 4/27 as the least number which u will get....
The expression is homogenous, so let a + b + c = 1 . Then we want to find the maximum of a b 2 + b c 2 + c a 2 which is a problem we have already encountered before.
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Nice one, Daniel! Same solution here.
Is there any proof without any assumptions?
Ah yes, I forgot I posted it a long time back under Maximizing a Cyclic Polynomial , and Michael posted a similar algebraic proof, along with calculus approaches.
I've always been amazed by this factorization, which is why I posted the problem again :)
The stronger ( a + b + c ) 3 a 2 b + b 2 c + c 2 a + a b c ≤ 2 7 4 is also true.
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Indeed. That also follows from the factorization, as we have the term 2 7 a b c .
what about a=0,b=0,c=1
How can you say that it's max value is 4/27??? Is there any way to prove it without assuming values of a,b and c??
Sir i feel its totally wrong (sorry if u feel its rude but i feel that only) this is my solution lets assume, a=b=c (as in most case when more than one variable is used the maximum is when all are equal, for instance sinx + cosx)
=>((a (a)^3)+(a (a)^3)+(a*(a)^3))/((a+a+a)^3)
=> (a^3+a^3+a^3)/(3a)^3
=> (3(a)^3)/9(a)^3
=>3/9
=>1/3
=>0.33
justification for my answer
a = 2 ; b = 2 ; c = 2
2³+2³+2³
(2*3)³
3*2³
9*2³
1
3
0.33
also i think .14 is less than .33 Hence your solution is wrong,
sir, please correct me if i am wrong i wrote this with a little experience i have maths, so please correct me if i am wrong.
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( 3 a ) 3 = 2 7 a 3 = 9 a 3
It is not true that the maximum/minimum must/can only occur when all values are equal, even if the expressions are symmetric or cyclic.
For example, subject to x + y = 2 , what is the minimium value of [ ( x − 2 ) ( y − 2 ) ] 2 ?
How can you say that maximum is 4 / 2 7
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I demonstrated that
1.
2
7
4
is an upper bound
2.
2
7
4
can be achieved by the values of
(
0
,
1
,
2
)
.
Hence, it is the maximum.
But due to the equality case, if a=b=c=1, then the maximum value of this would be 3.
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But a b 2 + b c 2 + c a 2 = 3 and ( a + b + c ) 3 = 2 7 , so your maximum is 9 1 . If you're referring to maximizing a b 2 + b c 2 + c a 2 under the condition that a + b + c = 1 , then note that 1 + 1 + 1 + = 1 . I hope this helps :D
Why must that be the maximum / equality case?
It is so well-known that a b 2 + b c 2 + c a 2 < = [ 4 ( a + b + c ) 3 ] / 2 7
hence the maximum is 4/27
Replace a=x^2,b=y^2,c=z^2 and x=rsinAsinB,y=rcosAsinB,z=rcosB,spherical coordinates and we know a+b+c=r r, so our function became f=(r r(sinA)^2 (sinB)^2 r^4 (cosA)^4 (sinB)^4+r r (cosA)^2 (sinB)^2 r^4 (cosB)^4+r r (cosB)^2 r^4 (sinA)^4 (sinB)^4)/(r^6) ,so f(A,B)=(sinB)^2 g(A,B) so we get max for (sinB)^2=1 and f become (cosA)^4-(cosA)^6 so it get max for one of the solutions of f'=0 which means sinA(cosA)^3 (-4+6(cosA)^2)=0 so we get max for (cos(A))^2=2/3,which means (sinA)^2=1/3 and our max is (2/3)^2-(2/3)^3=4/9-8/27=(12-8)/27=4/27...
We will show that
( a + b + c ) 3 a b 2 + b c 2 + c a 2 ≤ 2 7 4 ,
where equality occurs when a = 0 , b = 1 , c = 2 and cyclic permutations and multiples.
WLOG, let a be the minimum of the 3 variables. Observe that
= ≥ 4 ( a + b + c ) 3 − 2 7 ( a 2 b + b 2 c + c 2 a ) 9 a ( a 2 + b 2 + c 2 − a b − b c − c a ) + 2 7 a b c + ( 4 b + c − 5 a ) ( a + b − 2 c ) 2 0 + 0 + 0
Hence, the result follows.