If $a,b,c \geq 1$, then prove that
$4(abc+1) \geq (1+a)(1+b)(1+c)$
INEQUALITY PROBLEM
-
Prosenjit Basak
- Posts:53
- Joined:Wed Nov 28, 2012 12:48 pm
Yesterday is past, tomorrow is a mystery but today is a gift.
Re: INEQUALITY PROBLEM
Expand the RHS. The inequality will turn into $3abc+3\geq ab+bc+ca+a+b+c$.
Now we can again transform it into this: $0\geq ab-abc+c-1+bc-abc+a-1+ca-abc+b-1$
$\Rightarrow 0\geq (ab-1)(1-c)+(bc-1)(1-a)+(ca-1)(1-b)$
Which is absolutely true from the given condition.
Now we can again transform it into this: $0\geq ab-abc+c-1+bc-abc+a-1+ca-abc+b-1$
$\Rightarrow 0\geq (ab-1)(1-c)+(bc-1)(1-a)+(ca-1)(1-b)$
Which is absolutely true from the given condition.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$
-
Prosenjit Basak
- Posts:53
- Joined:Wed Nov 28, 2012 12:48 pm
Re: INEQUALITY PROBLEM
Hmm... I like the solution.It's quite easy.But can we solve it using AM-GM inequality? Actually I want that very solution.I gave this problem in a particular chapter regarding AM-GM. So I want that kind of solution. 
Yesterday is past, tomorrow is a mystery but today is a gift.