Home » Uncategorized » Inequality 51 (George Basdekis)

Inequality 51 (George Basdekis)


If \displaystyle a,b,c>0 then prove that

\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \frac{3}{\sqrt{2}}\cdot\sqrt{\frac{a^2+b^2+c^2}{a+b+c}}.


From the Cauchy – Schwarz Inequality we have that

\displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \left[\sum a(b+c)\right]\cdot\sum\frac{a}{(a+b)(b+c)}.

Expanding the RHS, we get that

\displaystyle \begin{aligned}\left(\sum\frac{a}{\sqrt{a+b}}\right)^2& \leq 2(ab+bc+ca)\cdot \frac{a^2+b^2+c^2+ab+bc+ca}{(a+b)(b+c)(c+a)}\\&\leq 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\end{aligned},

from the obvious \displaystyle ab+bc+ca\leq a^2+b^2+c^2.

Moreover, notice that the following beautiful Inequality holds, that is

\displaystyle 9(a+b)(b+c)(c+a)\geq 8(a+b+c)(ab+bc+ca).

Thus, we have that

\displaystyle 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}.

We have proved our Inequality, because \displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}, or

\displaystyle \sum\frac{a}{\sqrt{a+b}}\leq 3\sqrt{\frac{a^2+b^2+c^2}{2(a+b+c)}}, 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