Topics of studying inequalities $1$

Inequalities, by my part, is one of the most amazing topic of maths, the more you get through it,the more you get really amazed by it. So I will post some posts discussing some topics of inequalities. So, I will start with one of the most basic inequality i.e. Arithmetic mean-Geometric mean, often said as AM-GM inequality.Actually we will discuss QM-AM-GM-HM i.e. Quadratic mean,Arithmetic mean,Geometric mean,Harmonic mean. Now, these inequalities can be written as
$\sqrt{\frac{a^2+b^2}{2}}≥\frac{a+b}{2}≥\sqrt{ab}≥\frac{2}{\frac{1}{a}+\frac{1}{b}}$

This is just a general case. However the generalised formula can be written as

$\sqrt{\frac{a^2+b^2.....(n \; times)}{n}}≥\frac{a+b+c+....(n\;times)}{n}≥\sqrt[n]{a.b.....(n\;times)}≥\frac{n}{\frac{1}{a}+\frac{1}{b}+.....}$ The proof can be found here

Now, how can we apply these inequalities in solving various problems, Well, I can give some examples here to help you understand better.

$Example\;1$ let $a,b,c$ be positive real numbers. Now prove that $\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}≥2$

$Solution:$ Now, looking at what do we want to prove, we find that none of the means makes it simple, that the product of the terms cancel out some terms except some. So let us transform it a little.

$\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}+1≥3$

$=\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}≥3$ Now, we can easily prove this by AM GM inequality and we are done.

It is not always that we get direct transformation into means, Sometimes we have to transform them a bit, by breaking the inequalities.

$Example\;2$ Let $a,b,c$ be real numbers with sum $3$. Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c}≥ab+bc+ca$ (This question is from a Russian MO)

$Solution:$ We can check and see that directly applying means will be non beneficial. Because of the term on the right side. So let us try to remove it. We know that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$. Now, we see that we have $2(ab+bc+ca)$, So we can write as $2(\sqrt{a}+\sqrt{b}+\sqrt{c})≥2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)=9-(a^2+b^2+c^2)$ So, we have some nice terms. So we have to prove now that $(a^2+\sqrt{a}+\sqrt{a})+(b^2+\sqrt{b}+\sqrt{b})+(c^2+\sqrt{c}+\sqrt{c})≥9$ Now, applying AM-GM in each of the brackets , we get $(a^2+\sqrt{a}+\sqrt{a})+(b^2+\sqrt{b}+\sqrt{b})+(c^2+\sqrt{c}+\sqrt{c})≥3a+3b+3c=3(a+b+c)=9$ Hence, we have showed that $(a^2+\sqrt{a}+\sqrt{a})+(b^2+\sqrt{b}+\sqrt{b})+(c^2+\sqrt{c}+\sqrt{c})≥9$.

So, now we realize that the key technique in cracking this type of questions is to guess the correct use of means. I will provide some problems here for more practice:

Problem 1:. It is given that $a,b,c>0$ and $abc≤1$ Now prove that $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}≥a+b+c$

Problem 2:: Let $a,b,c$ be non negative numbers such that $a+b+c=2$. Prove that $2 ≥a^2b^2+b^2c^2+c^2a^2$

Problem 3:: let $a,b,c$ be positive real numbers such that $abc=1$ Prove that $\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{b+a}{\sqrt{c}}≥\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Here is the link for second part.