# Strenghtened Inequalities

All through ,i have oberved that (in mathlinks) the new inequalities that come up are alluring to think easy and with sufficient thought they exactly reverse our impressions. I observed that ,they mostly involved some well known expressions and had been devised to particularly make them look tough. Take for example the following inequality –

$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}+\frac{(b+c-a)(c+a-b)(a+b-c)}{abc}\ge 7$

or this –

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{3 (abc)^{\frac{1}{3}}}{2(a+b+c)}\ge 2$

The first i did get the solution using schur’ and it turned out to be not as tough as it looks. The second was easy too..but i didnt get a solution when i tried it some months back. The point is, i wanted to make such type of inequalities too and find some elegant solutions for them (to make up for my nature to get faulty with huge expressions :P). With such notions i thought , why not go from the basics.

What is the most basic inequality ? Ofcourse the idea of SOS ought to be the most basic but it certainly isnt what a beginner would learn. So, AM-GM has to be the top of this list.

I considered the inequality –

$a^2+b^2 \ge 2ab$

Hmm…. are you thinking ,what i am thinking?

Lets assume the truth of the following inequality –

$\frac{a^2+b^2}{ab} + \frac{(a+b)^2}{a^2+b^2} \ge 4$

It should be noted that the first term is greater than two and the second term less than 2 which made the inequality *not obvious*.

I checked some values of $a,b$ and saw it to be true and it is easily noted that the equality occurs for $a=b$. In no time, i did come up with a proof for the inequality –

$LHS = \frac{a^2+b^2}{ab} + \frac{2ab}{a^2+b^2} + 1$

By AM-GM for four positive reals,

$\frac{1}{4}\cdot\frac{a^2+b^2}{ab} + \frac{1}{4}\cdot\frac{a^2+b^2}{ab}+ \frac{ab}{a^2+b^2} + \frac{ab}{a^2+b^2} \ge 2$

Therfore –

$LHS = \frac{1}{2}\cdot\frac{a^2+b^2}{ab} + \frac{1}{4}\cdot\frac{a^2+b^2}{ab} + \frac{1}{4}\cdot\frac{a^2+b^2}{ab}+ \frac{ab}{a^2+b^2} + \frac{ab}{a^2+b^2} + 1 \ge 1 + 2 + 1 = 4$

Hurray! the first such inequality of mine! 😀

I hope i get better and tougher results then such namesake easy things.