# 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
5 years, 5 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

- 5 years, 1 month 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 - 5 years, 1 month ago

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

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

- 5 years, 1 month 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 - 5 years, 1 month ago

Yaaa agree here .

- 5 years, 1 month ago

Can it be done with AM GM HM Inequalities

- 5 years, 5 months ago

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

Staff - 5 years, 5 months ago