Home » Uncategorized » Inequality 36(Christos Patilas)

Inequality 36(Christos Patilas)

Problem:

If \displaystyle a,b,c are side lengths of a triangle, and \displaystyle E,R be its area and circumradius respectively, satisfying the equality \displaystyle a+b+c=3, prove that

\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(ER\right)^{\frac{1}{\sqrt{2}}}.

Solution:

Let \displaystyle E=\frac{abc}{4R}.

Then the given inequality is of the form

\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(\frac{abc}{4}\right)^{\frac{1}{\sqrt{2}}}.

Removing the square roots of \displaystyle 2 we get that

\displaystyle \left(\frac{3}{a^2+b^2+c^2}\right)^{\sqrt{2}}\geq abc.

Let us now introduce the \displaystyle u,v^2,w^3, that is

\displaystyle3u=a+b+c,3v^2=ab+bc+ca,w^3=abc.

Then the inequality takes the form

\displaystyle w^3\leq u^3\left(\frac{u^2}{3u^2-2v^2}\right)^{\sqrt{2}}.

Thus \displaystyle w^3 attains its maximum when \displaystyle 2 of the \displaystyle \left\{a,b,c\right\} are equal.

So, we need only need to prove the inequality for \displaystyle a=b, or we need to prove that for \displaystyle 2a+c=1 it holds that

\displaystyle \left(\frac{3}{2a^2+c^2}\right)^{\sqrt{2}}\geq a^2c,

which is true, 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