INEQUALITY'S BLOG

July 14, 2011

Problem 237 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 7:14 am

Let \displaystyle a, b, c\geq 0\&  \frac{1}{3a+1}+\frac{1}{3a+1}+\frac{1}{3a+1}=3. Prove that:

\displaystyle \frac{1}{3a+2}+\frac{1}{3b+2}+\frac{1}{3c+2}\geq \frac{1}{2}\biggl(\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\biggl)

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: