Given a, b, c are positive number, $abc=a+b+c$ prove $ab+4bc+9ac≥36$

Question

I know how to solve it using Cauchy-Schwarz Inequality (I find coefficient 1, 4, 9 for which the square of the sum of their square roots is exactly 36)?

But is there another way that involves less guessing and perhaps less complicated?

Answer 1

Here's an elementary-algebra-level solution . . .

Not instant, but it has the advantage of not requiring much sophistication.

Suppose $a,b,c > 0$ are such that $abc=a+b+c$.

Let $f=ab+4bc+9ac-36$.

Our goal is to show $f\ge 0$.

From $abc=a+b+c$, we get $c(ab-1)=a+b$, hence $ab > 1$, and $$c=\frac{a+b}{ab-1}$$ Replacing $c$ in $f$, we get $$f=\frac{(a^2+4)b^2-(24a)b+(9a^2+36)}{ab-1}$$ so it remains to show $g\ge 0$, where $$g=(a^2+4)b^2-(24a)b+(9a^2+36)$$ Regarding $g$ as a quadratic function of $b$, it suffices to show $D\le 0$, where $D$ is the discriminant $$D=(-24a)^2-4(a^2+4)(9a^2+36)$$ Expanding $D$ and then factoring yields $$D=-36(a+2)^2(a-2)^2$$ so $D\le 0$, which completes the proof.

Answer 2

Less complicated... unlikely.

Less guessing... Lagrange multipliers. Use the Lagrange multipliers to calcualte the minimum of $$f(a,b,c)=ab+4bc+9ac$$ given the condition that $$g(a,b,c)=abc-a-b-c=0.$$

That is, you get the Lagrangian function $L(a,b,c,\lambda)=ab+4bc+9ac-\lambda(abc-a-b-c)$, and the calculate the partial derivatives, and equate them to $0$, so you get the equations:

$$\frac{\partial L}{\partial a} = b+9c-\lambda bc+\lambda = 0\\ \frac{\partial L}{\partial b}=a+4c-\lambda ac+\lambda=0\\ \frac{\partial L}{\partial c}=4b+9a-\lambda ab+\lambda = 0\\ \frac{\partial L}{\partial \lambda} = a+b+c-abc=0$$

Solve the equations, and get the minimum value which is $\geq 36$.

Answer 3

Let $p=bc,q=ca,r=ab$.

$1=1/p+1/q+1/r$ and $p,q,r>0$.

Enough to show $p+4q+9r\ge 36$.

To make LHS zero-dimensional, multiply $1=1/p+1/q+1/r$.

Then we want to show $(p+4q+9r)(1/p+1/q+1/r)\ge 36$ but this is just direct consequence of Cauchy-Schwartz.

(Maybe you made a similar proof, but I want to emphasize that what you have to do is more obvious after this substitution, which is natural)

Answer 4

Let $a=2x$, $b=3y$ and $c=z$.

Thus, $$2x+3y+z=6xyz$$ and we need to prove that $$6xy+12yz+18xz\geq36$$ or $$xy+2yz+3xz\geq6\cdot\frac{6xyz}{2x+3y+z}$$ or $$(2x+3y+z)(xy+2yz+3xz)\geq36xyz,$$ which is true by AM-GM: $$(2x+3y+z)(xy+2yz+3xz)\geq6\sqrt[6]{x^2y^3z}\cdot6\sqrt[6]{xy(yz)^2(xz)^3}=36xyz.$$