Maximum value with inequality

Question

This is a problem that my friend and I are working on for olympiad training.

Let $a, b, c$ be real numbers in the interval $[0,1]$ that satisfy $ab+c \leq 1$. What is the maximum value of $a+b+c?$

I'm guessing the maximum is at $a,b=1$ and $c=0$, where we have $a+b+c=2$. As for proving this, I'm not sure how to proceed. Maybe AM-GM? Thanks for the help.

Answer 1

Let $S=a+b+c$ and fix $b$, then $$S\leq a+b +1-ab = a(1-b) +1+b =: f(a)$$

Since $b\leq 1$ we see that $f$ is increasing linear function, so $$f(a) \leq f(1) = 2$$

Since at $a=b=1$ and $c=0$ we see that $S_{\max}=2$.

Answer 2

1)$ab+c\le 1;$ $a, b, c \in [0,1];$

$d:=a+b+c$

The $2$ expressions are symmetric in $a$ and $b$. It follows that $a=b$ for $d_{max}$.

2)We want to find the maximum of

$d=2a+c,$ with $a^2+c \le 1$.

3)$a^2+d-2a \le 1;$

$(a-1)^2-1 +d \le 1;$

$d \le 2 - (a-1)^2.$

4)Maximal value for $d_{max}=2$ at $a=1;$

Hence $a=b=1$,and $c=0$.

Answer 3

Although other answers are valid, I offer this since it is simple. First note that since $a,b,c \in [0,1]$ the objective function $a+b+c$ is only increased when $c$ is increased. Therefore in the inequality $ab+c \le 1$ we can assume that in fact $c$ is large as possible for that constraint, i.e. that we have $c=1-ab.$

Hence the objective function becomes $a+b+(1-ab)=2-(1-a)(1-b).$ This is clearly at most $2$ and is actually $2$ for $a=b=1.$

Edit: Later I realized that to achieve the value $2$ requires only that one or both of $a,b$ be $1.$ So the max is obtained iff $(a,b,c)=(1,t,1-t)$ or $(t,1,1-t)$ where $0 \le t \le 1.$