# Prove $(2a + b + c)(a + 2b + c)(a + b + 2c) ≥ 64abc$ using the AM-GM method and establishing when inequality holds

Prove:

$$(2a + b + c)(a + 2b + c)(a + b + 2c) ≥ 64abc$$

for all $$a, b, c ≥ 0$$. Also, calculate/prove when equality holds.

To prove this, the first thing that came to mind was the Arithmetic Mean - Geometric Mean inequality (AM - GM) as $$64abc$$ can be written as $$8\sqrt{abc}$$. However, in this case, I would need to write the LHS in such a way where only a, b, and c remain and their sum is divided by two, which I'm a bit confused in doing. I also, tried subtracting the RHS from the LHS, but I ended up with a lot of numbers that weren't helpful for the proof. To establish when the equality holds, would it involve finding roots?

How would one go about proving this inequality with the AM - GM method? Any help would be extremely appreciated.

• Three identical answers all within one minute! A less elegant solution was posted here. Jun 10, 2019 at 6:33

$$2a+b+c = a+a+b+c\geq 4\sqrt[4]{a^2bc}$$ and similary for other two sums, so we get multiplying them: $$...\geq 64\sqrt[4]{a^4b^4c^4}=64abc$$

Equality hold iff all terms are equal in every inequality, so $$a=b=c$$

• Am-Gm for 4 terms can be proved with Am-Gm for 2 terms ${x+y\over 2}\geq \sqrt{xy}$ And here you can easly see that eqaulity holds iff $x=y$. Look here math.stackexchange.com/questions/64881/… Jun 10, 2019 at 7:37

$$2a+b+c=a+a+b+c\ge 4\sqrt[4]{a^2bc}$$ by AM/GM. Apply the same trick to each of the three brackets on the left.

Hint:

$$\dfrac{a+a+b+c}4\ge\sqrt[4]{a\cdot a bc}$$

the equality will occur if $$a=a=b=c$$

• Why is it that equality will hold when a = b = c? Jun 10, 2019 at 6:52
• Jun 10, 2019 at 6:59
• Is there a simpler way to prove a = b = c as opposed to the ones that the link provided? Jun 10, 2019 at 7:09
• @AlexanderB, $$a^4+b^4+c^4+d^4=(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2+4abcd\ge4abcd$$ if $ab-cd=a^2-b^2=c^2-d^2=0$ Jun 10, 2019 at 9:53