Home » Uncategorized » Inequality 29(George Basdekis)

Inequality 29(George Basdekis)

Problem:

If \displaystyle x,y,z,r>0 prove that

\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq 4r\sum_{cyc}x^3y.

We know that

\displaystyle 4r\sum_{cyc}x^3y\leq \left(r\sum_{cyc}x^3y+1\right)^2=r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1.

So, we only need to prove that

\displaystyle \frac{2}{3}r\cdot\left(\sum_{cyc}x^2\right)^{2}+r^{2}\sum_{cyc}x^4\sum_{cyc}x^2y^2+1\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y+1,

or

\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2+r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2+2r\sum_{cyc}x^3y.

But from Cauchy-Schwarz Inequality we deduce that

\displaystyle r^2\sum_{cyc}x^4\sum_{cyc}x^2y^2\geq r^2\left(\sum_{cyc}x^3y\right)^2.

So ,we only need to prove now that

\displaystyle \frac{2}{3}\cdot r\left(\sum_{cyc}x^2\right)^2\geq 2r\sum_{cyc}x^3y,

which is obviously true from Cirtoaje’s Inequality, Q.E.D.

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