Calculus and reasoning: If a^2+b^2+c^2+abc=4, show that

[jjm5119]

Using calculus and reasoning prove how, step by step, the following is true.

a, b, and c are >= 0

If a^2 + b^2 + c^2 +abc = 4 show that ab + bc + ac - abc <= 2

This is a calc 3 problem, vectors and planes in spaces, so i'm thinking i have to find the plane for the first equation and then show relationships to the other inequality. Basically, I'm pretty much lost and have no clue where to start. Any help is appreciated. Thanks.

 

[pka]

As you can see, the problem as stated is false.
\(\displaystyle \L \begin{array}{l}
a = \sqrt 2 \,,\,b = 1\,\& \,c = 1 \\
a^2 + b^2 + c^2 = 4 \\
ab + bc + ac - abc = \sqrt 2 + 1 > 2 \\
\end{array}\)

 

[jjm5119]

Made a mistake, first equation is suppose to be a^2 + b^2 + c^2 +abc = 4

 

[Deleted member 4993]

Duplicate post

http://mathgoodies.com/forums/topic.asp?TOPIC_ID=33044

 

[jjm5119]

am i not allowed to do that to try and get more answers?

 

[Deleted member 4993]

This is just to warn the tutors of existence of the other post - so that they do not duplicate their effort.