Algebraic inequality for positive reals $a,b,c$

Question

The problem is from a previous maths olympiad and the last step is to prove the inequality

$$4a^4bc + a^4c^2 + 9a^3bc^2 + 4a^3b^3 + 9a^2b^3c + a^2b^4 + 9ab^2c^3 + 4ab^4c + 4b^3c^3 + b^2c^4 +4a^3c^3 +4abc^4 \geq 24a^2b^2c^2 + 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$ for all $a,b,c > 0$.

I would very much appreciate if you could help me with how I can proceed from this step and finally solving the problem. Thank you

Answer 1

The hint:

You can use AM-GM and Rearrangement.

For example, by Rearrangement $$\sum_{cyc}(a^4bc-a^3b^2c)=abc\sum_{cyc}(a^3-a^2b)\geq0.$$ Indeed, the triples $(a^2,b^2,c^2)$ and $(a,b,c)$ have the same ordering.

Thus, $$\sum_{cyc}a^3=a^2\cdot a+b^2\cdot b+c^2\cdot c\geq a^2\cdot b+b^2\cdot c+c^2\cdot a=\sum_{cyc}a^2b.$$

Now, by AM-GM $$\sum_{cyc}(a^3b^3+a^3c^2b)\geq\sum_{cyc}2\sqrt{a^3b^3\cdot a^3c^2b}=2\sum_{cyc}a^3b^2c.$$ Thus, $$\sum_{cyc}(4a^4bc+3a^3b^3+3a^3c^2b)\geq10\sum_{cyc}a^3b^2c.$$ Id est, it's enough to prove that $$\sum_{cyc}(a^4c^2+a^3b^3+6a^3c^2b)\geq24a^2b^2c^2,$$ which is AM-GM again.

Answer 2

Hint: Use AM-GM inequality, like:

$$8a^3bc^2 + 8a^2b^3c + 8ab^2c^3\geq 24a^2b^2c^2$$ and then you are left to prove:

$$4a^4bc + 4b^3c^3+a^4c^2 + a^3bc^2 + 4a^3b^3 + a^2b^3c + a^2b^4 + ab^2c^3 + 4ab^4c + b^2c^4 +4abc^4 \geq 10a^3b^2c + 10a^2bc^3 + 10ab^3c^2$$