Home » Uncategorized » Inequality 21(Vasile Cirtoaje)

Inequality 21(Vasile Cirtoaje)


Let \displaystyle x,y,z be non-negative numbers. If \displaystyle 0\leq r\leq \sqrt{2}, prove that

\displaystyle \sqrt{x^4+y^4+z^4}+r\sqrt{x^2y^2+y^2z^2+z^2x^2}\geq (1+r)\sqrt{x^3y+y^3z+z^3x}.


If we square both sides we get

\displaystyle \sum_{cyc}x^4+r^2\sum_{cyc}x^2y^2+2r\sqrt{\sum_{cyc}x^4\sum_{cyc}x^2y^2}\geq r^2\sum_{cyc}x^3y+\sum_{cyc}x^3y+2r\sum_{cyc}x^3y.

Now from Cauchy-Schwartz inequality we know that \displaystyle 2r\sqrt{\sum_{cyc}x^4\sum_{cyc}x^2y^2}\geq 2r\sum_{cyc}x^3y.

So, it suffices to prove that

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

For \displaystyle r=0 the inequality is true. So, we only need to prove it for \displaystyle 0<r\leq \sqrt{2}.

Rewrite the inequality in the form

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

We know that \displaystyle \sum_{cyc}x^4-\sum_{cyc}x^3y\geq 0 so, it is enough to prove it for \displaystyle r=\sqrt{2}.

For \displaystyle r=\sqrt{2} we have that

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

which reduces to

\displaystyle \left(\sum_{cyc}x^2\right)^2\geq 3\sum_{cyc}x^3y, which is a well known inequality and we have proved it here: Well-known inequality, Q.E.D.


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