work 027 – mathlinks user CSS-MU

Problem

Let a,b,c be nonnegative numbers, no two of which are zero. Prove that:
\frac {ab + bc - ca}{c^2 + a^2} + \frac {bc + ca - ab}{a^2 + b^2} + \frac {ca + ab - bc}{b^2 + c^2} \geqslant \frac {3}{2}

Proof

It is equivalent to –
\sum (a - b)^2 \cdot (\frac {\sum ab - a^2)}{(c^2 + a^2)(c^2 + b^2)} \ge 0

Therefore ,
S_a = (\sum ab - a^2)(b^2 + c^2)
S_b = (\sum ab - b^2)(a^2 + c^2)
S_c = (\sum ab - c^2)(a^2 + b^2)

Let , a\ge b\ge c
So, S_c \ge S_b \ge S_a
and S_c , S_b \ge 0

also note that –
S_a + S_b = (\sum ab - a^2)(b^2 + c^2) + (\sum ab - b^2)(a^2 + c^2)

= b^2(\sum ab - a^2) + a^2(\sum ab - b^2) + c^2(2ab - a^2 - b^2 + 2c(a + b))

= (ab - c^2)(a - b)^2 + c(a + b)(a^2 + b^2) + 2c^3(a + b) \ge 0
thus the inequality is proved.

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: