Evaluating $\max(ab+bc+ac)$

Question

Let a,b,c be real numbers such that $a+2b+c=4$. What is the value of $\max(ab+bc+ac)$

My attempt:

Squaring both the sides:

$a^2 +4b^2+c^2+2ac+4bc+4ab=16$

Then I tried factoring after bringing 16 to LHS but couldn't. It's not even a quadratic in one variable or else I could have directly found the maximum.

How do I proceed?

Answer 1

From the given condition $\,2b=4-a-c\,$, then:

$$\require{cancel} \begin{align} 2(ab+bc+ca) &= 2b(a+c)+2ac \\ &= (a+c)(4-a-c)+2ac= \\ &= 4a - a^2 - \bcancel{ac} + 4c - \bcancel{ac} - c^2 + \bcancel{2ac} = \\ &= \color{red}{8} -a^2 + 4a \color{red}{-4} - c^2 +4c \color{red}{-4} = \\ &= 8 - (a-2)^2 - (c-2)^2 \\ &\le 8 \end{align} $$

Equality is attained for $a=c=2\,$, so the upper bound is in fact a maximum.

Answer 2

Lagrange multipliers will give the solution \begin{eqnarray*} L=ab+bc+ca+ \lambda (a+2b+c-4) \end{eqnarray*} Differentiating gives \begin{eqnarray*} b+c+ \lambda =0 \\ c+a +2\lambda =0 \\ a+b+ \lambda =0 \\ \end{eqnarray*} Add these equation together and substitute to get $ \lambda = \frac{b-4}{2}$ now substitute this back into the equations above and we get $b=0, a=c=2$. So the maximum value is $\color{blue}{4}$.

Edit : Ooops ... it is a precalculus question ... \begin{eqnarray*} L=ab+bc+ca=ab+(a+b)(4-a-2b)=-a^2-2ab-2b^2+4(a+b) = \\ \underbrace{-\frac{1}{2}(2b+a-2)^2}_{2b+a=2}-\underbrace{\frac{1}{2}(a-2)^2}_{a=2}+\color{blue}{4} \end{eqnarray*}

Answer 3

Note that $4xy \leq (x+y)^2$ for all $x$, $y$.

You have $$ab+bc+ca = b(a+c) + ac \leq b(a+c) + \frac{1}{4}(a+c)^2 = (a+c) \frac{a+c+4b}{4} = \frac{1}{4}(a+c)(8-(a+c)) \leq \frac{1}{4}\frac{1}{4}8^2 = 4.$$

Equality is happen when $a=c$ and $a+c = 8-(a+c)$, or $a+c=4$.