4 dimensional inequality

Given non-negative reals such that a+b+c+d=4 a + b + c + d = 4 , prove that

a2bc+b2cd+c2da+d2ab4. a^2bc + b^2 cd + c^2 da + d^2 ab \leq 4.

Hint: Solve this problem.

Note by Calvin Lin
5 years, 9 months ago

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Can it be done with AM GM HM Inequalities

Satvik Choudhary - 5 years, 9 months ago

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It can be done with AM-GM, but not in a standard way. Note that it has "strange equality conditions".

Calvin Lin Staff - 5 years, 9 months ago

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Let f(a,b,c,d)=a2bc+b2cd+c2da+d2ab+k(a+b+c+d4)f(a,b,c,d)=a^2bc + b^2cd + c^2da + d^2ab + k(a+b+c+d-4) (say this as equation (1))(1))

Solving these five equations fa=0\frac{\partial f}{\partial a} = 0 fb=0\frac{\partial f}{\partial b} = 0 fc=0\frac{\partial f}{\partial c} = 0 fd=0\frac{\partial f}{\partial d} = 0 a+b+c+d=4a+b+c+d=4

we get a=b=c=d=1,k=4a=b=c=d=1 , k=-4

substituting this back in equation 11 , we get fmax(a,b,c)=1+1+1+14(0)=4f_{max}(a,b,c)=1+1+1+1-4(0) = 4


@Calvin Lin sir

Aman Rajput - 5 years, 6 months ago

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The goal is to not use calculus.

Furthermore, note that (2,1,1,0) (2, 1, 1, 0 ) is another equality condition, which is how I know that you didn't do the Lagrangian properly. There is a priori no reason why the answer must be symmetric.

Calvin Lin Staff - 5 years, 6 months ago

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Note again that ... case (2,1,1,0) doesnt satisfy the fourth equation

at this case fd0\frac{\partial f}{\partial d} \neq 0

Aman Rajput - 5 years, 5 months ago

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@Aman Rajput Once again, you are not considering the boundary condition restraints. There is no argument that "f(2, 1, 1, 0) satisfies the equality condition", and so if it doesn't appear in your solution you have to ask yourself what is the mistake that you made.

If you want to apply a theorem, make sure you use it exactly and completely, and that you check all of the necessary conditions. E.g. Do not apply Arithmetic Mean - Geometric Mean on negative numbers.

Calvin Lin Staff - 5 years, 5 months ago

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@Calvin Lin Yaaa agree here .

Aman Rajput - 5 years, 5 months ago

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