work 17 – CRUX

Prove that for a,b,c sides of triangle the following inequality hold good-

3 \ge \sum \frac{a(b+c)}{a^2+bc}

Proof

3 - \sum \frac{a(b+c)}{a^2+bc} = \sum \frac{(a-b)(a-c)}{a^2+bc}

So we have to prove that –

\sum \frac{(a-b)(a-c)}{a^2+bc} \ge 0

Note that –

b^2+ac \ge c^2+ab as a,b,c are sidelengths of a triangle

Therefore.

\frac{1}{c^2+ab} \ge \frac{1}{b^2+ac}

thus by Vornicu-Schur the inequality is true

Advertisements

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

%d bloggers like this: