What are the most popular techniques of proving inequalities?

There are also:

1. TL (The Tangent Line Method)

2. SOS (Sum of Squares)

3. Schur (Inequalities like the Schur's inequalities)

4. SS (SOS-Schur metod)

5. Muirhead (Muirhead inequalities)

6. Karamata

7. Rearragement

8. Chebyshov

9. Holder

10. Bacteria (undefined coefficients method)

11. Discriminant

12. The $uvw$'s technique (it's not the $uvw$ method)

13. Minkowski (triangle inequality)

14. Bernoulli

15. LM ( Lagrange Multipliers method)

16. EV (The Vasc's Equal Variables Method)

17. RCF, LCF (the Vasc's Right Convex Function Method and Left Convex Function Method)

18. Zhaobin (about half convex- half concave function)

19. prR (for the geometric inequalities proofs)

20. BW (the Buffalo Way method)

and more.

A very good book it's "Algebraic inequalities, old and new methods", Gil, 2006 by Vasile Cirtoaje.

Also, thanks to Jose Brox, there is the beautiful Vasc's last book about inequalities, which he published in 2015.

The example, how the Bacteria method helps to find a proof of the very hard inequality.

Let $a\geq0$, $b\geq0$ and $c\geq0$ such that $a+b+c=3$. Prove that: $$\frac{1}{8+a^2b}+\frac{1}{8+b^2c}+\frac{1}{8+c^2a}\geq\frac{1}{3}.$$ This inequality is ninth degree.

Now, we'll reduce this degree.

By C-S $$\sum_{cyc}\frac{1}{8+a^2b}=\sum_{cyc}\frac{(a+kb+mc)^2}{(a+kb+mc)^2(8+a^2b)}\geq\frac{(1+k+m)^2(a+b+c)^2}{\sum\limits_{cyc}(a+kb+mc)^2(8+a^2b)}.$$ We'll choose values of $k$ and $m$ ( they are our bacteria) such that the inequality $$\frac{(1+k+m)^2(a+b+c)^2}{\sum\limits_{cyc}(a+kb+mc)^2(8+a^2b)}\geq\frac{1}{3}$$ would be true.

Since the equality in the starting inequality occurs for $a=b=c=1$ and again for

$(a,b,c)=(1,0,2)$ and for cyclic permutations of the last, we get the following system: $$\frac{a+kb+mc}{(a+kb+mc)^2(8+a^2b)}=\frac{b+kc+ma}{(b+kc+ma)^2(8+b^2c)}= \frac{c+ka+mb}{(c+ka+mb)^2(8+c^2a)}$$ or $$\frac{1}{(a+kb+mc)(8+a^2b)}=\frac{1}{(b+kc+ma)(8+b^2c)}= \frac{1}{(c+ka+mb)(8+c^2a)},$$ which is obviously true for $a=b=c=1$.

But for $(a,b,c)=(1,0,2)$ we obtain: $$\frac{1}{8(1+2m)}=\frac{1}{8(2k+m)}= \frac{1}{12(2+k)},$$ which gives $k=\frac{8}{5}$, $m=\frac{11}{5}$ and we can write: $$\sum_{cyc}\frac{1}{8+a^2b}=\sum_{cyc}\frac{(5a+8b+11c)^2}{(5a+8b+11c)^2(8+a^2b)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(5a+8b+11c)\right)^2}{\sum\limits_{cyc}(5a+8b+11c)^2(8+a^2b)}=\frac{576(a+b+c)^2}{\sum\limits_{cyc}(5a+8b+11c)^2(8+a^2b)}.$$ Id est, it's enough to prove that: $$\frac{576(a+b+c)^2}{\sum\limits_{cyc}(5a+8b+11c)^2(8+a^2b)}\geq\frac{1}{3},$$ which is fifth degree and the rest is smooth.

The book Inequalities by Por G. H. Hardy, J. E. Littlewood, G. Pólya.