An inequality for the christmas eve !!

I usually categorize inequalities into 4 types (which is a rather weird kind of classification) keeping in mind the problem solving strategies available:

TYPE 1 : <summation> $sign$ <summation>
TYPE 2 : <summation> $sign$ <product>
TYPE 3 : <product> $sign$ <summation>
TYPE 4 : <product> $sign$ <product>

 "<summation>" denotes expressions majorly involving sums and differences which may be symmetric or asymmetric like $ \sum ab$ or $\sum \frac{ab}{c-1}$ or $a+b^2+c^3$ or $\sqrt{\frac{a^2+bc}{b+c}}$ or more complicated stuff.
"<product>" denotes symmetric or asymmetric expressions involving products like $\prod (a-1)$ or $abcd$ or $(\sum ab)^2 (\sum a)(abc)$ etc ...
"$sign$" may be $\ge$or$\le$or$<$or$>$.

I'll basically be talking of algebraic inequalities in this article while some of the rules are inherited by geometric inequalities too but they need a seperate discussion (topic of the next post).


TYPE 1, 2 engulf a very large portion of the whole subject and some of the techniques and theorems are easily applicable to them.The statements of most of the theorems in inequalities belong to these categories (like Weighted means, Cauchy-Schwarz, Tchebycheff's, Jensen's, Muirhead's, ...)
TYPE 4 can be converted to TYPE 1 by suitable transformations and substitutions (simplest of the methods being applying log on both sides) or solved using Muirhead's inequality.
TYPE 3 is a difficult type. My experience shows me that Schur's inequality is a strong technique to prove problems of this category (Muirhead's is also sometimes helpful). Most of the time we have to rely on some ingenuine and innovative methods of substitution(s) and transformations which differ from problem to problem which either solve the problem directly or reduce it to that of TYPE 1 or 2.

Here I will present a small new technique to prove certain symmetric (sometimes asymmetric also) inequalities of TYPE 4 based on Lagrange's identity which states :

$(a^2 +nb^2)(c^2+nd^2)=(ac-nbd)^2+n(ad+bc)^2=(ac+nbd)^2+n(ad-bc)^2$

Let me demonstrate with an example :

Prove the following :

$(a^2+2)(b^2+2)(c^2+2) \ge 9 \sum ab$

Proof:

$(a^2+2)(b^2+2)(c^2+2) = ((\sqrt{2}a+\sqrt{2}b)^2+(ab-2)^2)(c^2+2)$

                      $= (2a+2b+2c-abc)^2+(\sqrt{2}ab+\sqrt{2}bc+\sqrt{2}ca-2\sqrt{2})^2 $

                      $=(2\sum a-abc)^2+2(\sum ab -2)^2 $

(we have basically used the identity to put the expression as a combination of elementary symmetric functions)


Let $ \sum a =p,\; \sum ab=q,\,  abc=r $. Then we have the following basic inequalities relating elementary symmetric functions of $a, \; b, \; c$ :

$p \ge \sqrt{3q} \ge (27r)^\frac{1}{3}\\ q^2 \ge 3pr $

What we need to prove is :

$4p^2+r^2-4pr+2q^2+8 \ge 17q$

We now have to carefully break down the terms on the left side to obtain the result using the above basic inequalities. Looking at the equality case always helps! Equality clearly holds when $a=b=c=1$ i.e. $p=3, \; q=3, \; r=1$. The terms which are creating a 'problem' are $r^2$ and $8$. Notice how we split the terms with an aim to apply AM-GM inequality :

$4p^2+r^2-4pr+2q^2+8 = (\frac{p^2}{9}+r^2)+\frac{35p^2}{9}-4pr+(\frac{8q^2}{9}+8)+\frac{10q^2}{9}$

                   $\ge \frac{2pr}{3}+\frac{35p^2}{9} -4pr+\frac{16q}{3}+\frac{10q^2}{9}$

                   $=\frac{35p^2}{9}+\frac{16q}{3}+\frac{10q^2}{9}-\frac{10pr}{3}$

                   $\ge \frac{35p^2}{9}+\frac{16q}{3}$

                   $\ge \frac{35q}{3}+\frac{16q}{3}$

                   $=17q$
Q.E.D.

Any other solutions for the above inequality are welcome :-)

COMING UP NEXT : On Geometric Inequalities....

A Merry Christmas to all !!

AAKANSH