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# 4 dimensional inequality

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

$a^2bc + b^2 cd + c^2 da + d^2 ab \leq 4.$

Hint: Solve this problem.

Note by Calvin Lin
2 years ago

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Comment deleted Sep 16, 2015

Hm, how did you get the first inequality? It doesn't seem true with $$a = 2, b = c = 1, d = 0$$. The LHS is 2, while the RHS is 4. Staff · 1 year, 8 months ago

Can it be done with AM GM HM Inequalities · 2 years ago

It can be done with AM-GM, but not in a standard way. Note that it has "strange equality conditions". Staff · 2 years ago

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

Solving these five equations $\frac{\partial f}{\partial a} = 0$ $\frac{\partial f}{\partial b} = 0$ $\frac{\partial f}{\partial c} = 0$ $\frac{\partial f}{\partial d} = 0$ $a+b+c+d=4$

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

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

$$\textbf{Q.E.D}$$

@Calvin Lin sir · 1 year, 8 months ago

The goal is to not use calculus.

Furthermore, note that $$(2, 1, 1, 0 )$$ 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. Staff · 1 year, 8 months ago

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

at this case $$\frac{\partial f}{\partial d} \neq 0$$ · 1 year, 8 months ago

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. Staff · 1 year, 8 months ago