Home » Uncategorized » Inequality 7(Christos Patilas)

Inequality 7(Christos Patilas)

Problem:

Let \displaystyle a,b,c\in \mathbb{R^{*}} and \displaystyle \sqrt{a^4+b^4} \geq \sqrt{a^4+c^4}+\sqrt{b^4+c^4} and \displaystyle |ab| \geq |ac|+|bc| . Prove that

\displaystyle \left[\frac{\sqrt{a^4+b^4}-\sqrt{b^4+c^4}-\sqrt{c^4+a^4}}{2}\right]^{2}+\frac{\left(\left|ab\right|-\left|bc\right|-\left|ca\right|\right)^{2}}{2}\geq c^{4} .

Solution:

We will use the following lemma:

For \displaystyle 0<p<1 it holds that

\displaystyle \left(a^{p}_{1}-a^{p}_{2}-a^{p}_{3}\right)^{\frac{1}{p}}+\left(b^{p}_{1}-b^{p}_{2}-b^{p}_{3}\right)^{\frac{1}{p}}\geq \left[\left(a_{1}+b_{1}\right)^{p}-\left(a_{2}+b_{2}\right)^{p}-\left(a_{3}+b_{3}\right)^{p}\right]^{\frac{1}{p}}.

So, substituting  \displaystyle a_{1}=a^4+b^4,a_{2}=b^4+c^4,a_{3}+c^4+a^4\wedge b_{1}=2a^2b^2,b_{2}=2b^2c^2,b_{3}=2c^2a^2\wedge p=\frac{1}{2} we get that

\displaystyle \left(\sqrt{a^4+b^4}-\sqrt{b^4+c^4}-\sqrt{c^4+a^4}\right)^{2}+\left[\sqrt{2}\left(\left|ab\right|-\left|bc\right|-\left|ca\right|\right)\right]^{2}\geq \left(\sqrt{(a^2+b^2)^2}-\sqrt{(b^2+c^2)^2}-\sqrt{(c^2+a^2)^2}\right)^{2}.

Dividing the last relation with \displaystyle 4 we have that

\displaystyle \begin{aligned}\left[\frac{\sqrt{a^4+b^4}-\sqrt{b^4+c^4}-\sqrt{c^4+a^4}}{2}\right]^{2}+\frac{\left(\left|ab\right|-\left|bc\right|-\left|ca\right|\right)^{2}}{2}&\geq \frac{\left(a^2+b^2-b^2-c^2-c^2-a^2\right)^{2}}{4}\\&=c^{4}\end{aligned}, 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