Home » Uncategorized » Inequality 9(Christos Patilas)

Inequality 9(Christos Patilas)


If \displaystyle a_{1},a_{2},a_{3} are the positive real roots of the equation \displaystyle 4x^3-kx^2+mx-9=0 prove that

\displaystyle k\geq 4\sqrt[3]{\sum_{cyc}\left(a_{1}\sqrt{a_{2}+a_{3}}\right)+3\prod_{cyc}(a_{1}+a_{2})}.


Let us divide both sides by \displaystyle 4 and then cube them.

We acquire

\displaystyle \left(\frac{k}{4}\right)^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}).

But from Viete’s relations we have that

\displaystyle \frac{k}{4}=a_{1}+a_{2}+a_{3} and \displaystyle a_{1}a_{2}a_{3}=\frac{9}{4}.

So our inequality transforms into

\displaystyle (a_{1}+a_{2}+a_{3})^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}),


\displaystyle \sum_{cyc} a^{3}_{1}+3\prod_{cyc}(a_{1}+a_{2}) \geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}).

So, it suffices to prove that

\displaystyle \sum_{cyc} a^{3}_{1}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}.

But the last inequality holds because

\displaystyle a^{3}_{1}+a^{3}_{2}+a^{3}_{3}\geq a_{1}a_{2}(a_{1}+a_{2})+a^{3}_{3}\geq 2\sqrt{a_{1}a_{2}a_{3}\cdot a^{2}_{3}(a_{1}+a_{2})}=3a_{3}\sqrt{a_{1}+a_{2}}.

Adding up the \displaystyle 3 cyclic relations we come to the desired 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 )

Google+ photo

You are commenting using your Google+ 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 )


Connecting to %s