Prove that $2a + 2ab + abc \leq 18$ when $a + b + c = 5$ where $a, b, c \in \mathbb{R}$ without calculus.

Question

I have already had help with proving the conditional statement on this post.

Apparently, it includes the following trick from this website, and it says it works for any cubic polynomial function.

However, it does not seem to be true for any cubic polynomial function such as $P = x^3 + 3x$. This gives me some doubt of using a similar approach shown on this or this.

I was wondering if there is another way to prove my statement without calculus or the previous methods from this post and this post.

If so, please show another step-by-step solution. Thank you!

As you can see, I was trying the AM-GM inequality before, but I got stuck.

Answer 1

I'll just answer the specific question of $P(x)=x^3 + 3x$ here.

First we prove that $P(x) > P(y)$ whenever $x>y$:

$$ P(x)-P(y) = x^3-y^3 + 3x-3y = (x-y)(x^2+xy+y^2+3)$$

But $x^2 + xy + y^2 = x^2( k^2 + k + 1) \ge 0$, where $k = y/x$. The inequality follows from the fact that $k^2 + k + 1 \ge k^2 + k + \frac14 = (k+\frac12)^2 \ge 0$. So the RHS is positive.

So if you restrict $a \le x \le b$, then $P(b)$ will be the maximum, because since $x \le b$, $P(x) \le P(b)$ by our proof above. The minimum is similarly $P(a)$.

Answer 2

By AM-GM $$2a+2ab+abc=a(2+b(2+c))\leq a\left(2+\left(\frac{b+2+c}{2}\right)^2\right)=a\left(2+\left(\frac{7-a}{2}\right)^2\right)\leq18,$$ where the last inequality it's $$a(57-14a+a^2)\leq72$$ or $$a^3-14a^2+57a-72\leq0$$ or $$a^3-6a^2+9a-8a^2+48a-72\leq0$$ or $$(8-a)(a-3)^2\geq0,$$ which is obvious.

Another way.

We need to prove that: $$\frac{2a(a+b+c)^2}{25}+\frac{2ab(a+b+c)}{5}+abc\leq\frac{18(a+b+c)^3}{125}$$ or $$18c^3+(44a+54b)c^2+(34a^2-87ab+54b^2)c+2(4a^3-8a^2b-3ab^2+9b^3)\geq0,$$ for which it's enough to prove that: $$2(22a+27b)c^2+(2a-3b)(17a-18b)c+2(2a-3b)^2(a+b)\geq0,$$ for which it's enough to prove that $$(17a-18b)^2-16(a+b)(22a+27b)\leq0$$ or $$63a^2+1396ab+108b^2\geq0.$$