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.

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## Comments

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TopNewestCan it be done with AM GM HM Inequalities

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

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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

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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.

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

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

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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.

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