work 011 – Kvant No.4 1990

Problem

Prove the following inequality –

ax + by +cz + \sqrt{(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)} \ge \frac{2}{3} \cdot A

where

A = (a+b+c)(x+y+z)

for Reals a,b,c,x,y,z

My Solution

The inequality is cyclic. So assume that b,y are the middle elements of (a,b,c) ; (x,y,z) respectively.

By Cauchy-Schwarz Inequality,

\sqrt{(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)} \ge az + by + cx

Therefore it is sufficient to prove that –

(a + c)(x + z) + 2by \ge RHS

\Leftrightarrow (x + z - 2y)(a + c - 2b) \ge 0

which is true from the assumption .

Hence the inequality is proved with equality for a=b=c and x=y=z

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: