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.

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## Comments

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TopNewestCan you add this to the Brilliant Wiki pages for relevant skills in Classical Inequalities? Thanks!

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I am glad to add it, Sir. But I have a doubt. Do I have to just copy this note in latex and click on "Add a Post" in the practice map or something else ........... ?

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If you click on Classical Inequalities, that will bring up the skill tree. If you click on a specific skill, that will bring you to the Wiki page, and you can edit it directly.

For example, the first skill of Trivial Inequality, has the corresponding WIki page

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I hope that is Just copying the note. I will do it.But sir, can you plz check my latex error in second part. It looks OK when I checked it on Stackexchange.. I will add it too and will try to write a more few notes

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Awesome notes.............

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Counterexample for problem 1: \(a = 100, b = \frac{1}{10000}, c = 100\) gives \(2.01 \geq 200.0001\) =_="

Is that a typo or stuffs?

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Oh, yes, I had mad a small mistake while writing, I will edit it now

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I don' understand example 1, can you explain how you prove it after adding \(1\) to it

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OK, @Daniel Lim , Try to see the defination of AM-GM. While taking the GM of the numbers, we take the product of those numbers. But It is difficult to deal with Some complicated products withing roots. So, if we can make the product of the numbers equal to 1, then it would be easy enough. Now, looking at the original problem, you can see that if you apply AM-GM, then it becomes \(\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}≥3\sqrt[3]{\frac{b}{b+c}}\).

Now, at this result, we are struck, because we cannot do any direct application from here.But, if we can transform \(\frac{b}{b+c}\) Magically into \(1\), then our work becomes easy,as it would directly conclude that

\(\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}≥3\)

SO, how do we do this task. Now, notice that when we add 1 to our original inequality, we notice that

\(\frac{c}{a}+\frac{a}{b+c}+\frac{b}{c}+1≥2+1\) Which is \(\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}≥3\) Now, when we apply AM-GM Here, we get

\(\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}≥3\sqrt[3]{\frac{c}{a}.\frac{a}{b+c}.\frac{b+c}{c}}=3\)

Which is what we want to show. I hope you have understood it now. If there is still any doubt, the please ask..

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Ok, I think I know what's my problem, I don' understand what does "apply AM-GM" mean, therefore I can't understand your solution

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I also can't understand example 2

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Can you tell from which step you didn't understood ?

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Thats a lotta stuff dere...can any1 explain dis stuff???

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Which Part Actually, Can you tell more clearly ?

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Wow! Very helpful! :)

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Great Note! Keep up the good work!

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