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.

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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– Calvin Lin Staff · 1 year, 8 months ago

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.Log in to reply

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

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– Calvin Lin Staff · 2 years ago

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

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 – Aman Rajput · 1 year, 8 months ago

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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. – Calvin Lin Staff · 1 year, 8 months ago

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at this case \(\frac{\partial f}{\partial d} \neq 0\) – Aman Rajput · 1 year, 8 months ago

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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. – Calvin Lin Staff · 1 year, 8 months ago

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– Aman Rajput · 1 year, 8 months ago

Yaaa agree here .Log in to reply