I am stuck with this problem:
Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$
I tried to use: $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc \implies P=\frac{1+abc}{a+b+c}$ but I am stuck. Can anyone help me? Thank you!
I'm just 14 years old, so don't use derivative or something like that (my brother uses it but I cant understand).
On
For any numbers $x, y, z > 0$, the AM-GM inequality gives $$ \frac{x + y + z}{3} \ge \sqrt[3]{xyz}, $$ which is equivalent to $$ (x + y + z)^3 \ge 27 xyz, $$ where equality holds if and only if $x = y = z$.
Setting $x = a + b$, $y = b + c$, $z = c + a$, we get $x + y + z = 2a + 2b + 2c$, so $$ (2a + 2b + 2c)^3 \ge 27 \cdot (a + b)(b + c)(c + a) = 27, $$ so $$ 2a + 2b + 2c \ge \sqrt[3]{27} = 3, $$ giving $$ a + b + c \ge \frac{3}{2}, $$ where equality holds if and only if $a + b = b + c = c + a$; that is, $a = b = c = \frac{1}{2}$.
Setting $x = a$, $y = b$, $z = c$, we get $$ abc \le \frac{1}{27} \cdot (a + b + c)^3, $$ where equality holds if and only if $a = b = c$.
Therefore, as you have calculated, $$ P = \frac{1 + abc}{a + b + c} \le \frac{1 + \frac{1}{27} (a + b + c)^3}{a + b + c} = \frac{(a + b + c)^2}{27} + \frac{1}{a + b + c} $$ where equality holds if and only if $a = b = c = \frac{1}{2}$.
Let $$ f(x) = \frac{x^2}{27} + \frac{1}{x}. $$ To determine the minimum point of $f$ without derivatives, we apply the AM-GM inequality again. When $x > 0$, $$ f(x) = \frac{x^2}{27} + \frac{1}{2x} + \frac{1}{2x} \ge 3 \cdot \sqrt[3]{\frac{x^2}{27} \cdot \frac{1}{2x} \cdot \frac{1}{2x}} = \frac{1}{\sqrt[3]{4}}, $$ where equality holds if and only if $\frac{x^2}{27} = \frac{1}{2x} = \frac{1}{2x}$; that is, $x = \frac{3}{\sqrt[3]{2}}$.
For any $0 < x < y$, $$ f(x) - f(y) = \frac{x^2 - y^2}{27} + \frac{1}{x} - \frac{1}{y} = \frac{(x+y)(x-y)}{27} - \frac{x - y}{xy} = \frac{(x-y)(x^2 y + x y^2 - 27)}{27xy}. $$ When $0 < x < y \le \frac{3}{\sqrt[3]{2}}$, $$ x^2 y + x y^2 - 27 < \frac{27}{2} + \frac{27}{2} - 27 = 0, $$ so $f(x) > f(y)$. Similarly, when $\frac{3}{\sqrt[3]{2}} \le x < y$ $f(x) < f(y)$. In other words, $f$ is decreasing on $(0, \frac{3}{\sqrt[3]{2}})$ and increasing on $(\frac{3}{\sqrt[3]{2}}, +\infty)$.
Now we consider two cases.
(1) If $\frac{3}{2} \le a + b + c \le 3$ (note that $\frac{3}{2} < \frac{3}{\sqrt[3]{2}} < 3$), then $$ P \le f(a + b + c) \le \max\{f(\tfrac{3}{2}), f(3)\} = \max\{\tfrac{3}{4}, \tfrac{2}{3}\} = \tfrac{3}{4}, $$ where equality holds if and only if $a + b + c = \frac{3}{2}$, which in turn holds if and only if $a = b = c = \frac{3}{2}$.
(2) If $a + b + c > 3$, then at least one of $a$, $b$, and $c$ is greater than $1$. Assume without loss of generality that $c > 1$. Since $$ a + b = \frac{1}{(a + c)(b + c)} < \frac{1}{c^2}, $$ we have $$ c > 3 - (a + b) > 3 - \frac{1}{c^2} > 2, $$ so $a + b < \frac{1}{4}$. Hence, $$ ab \le \biggl(\frac{a + b}{2}\biggr)^2 < \frac{1}{64} $$ by the AM-GM inequality. Therefore, $$ P = \frac{1}{a + b + c} + \frac{abc}{a + b + c} < \frac{2}{3} + \frac{\frac{1}{64} c}{c} = \frac{2}{3} + \frac{1}{64} < \frac{3}{4} $$ (note that $a + b + c > c$).
Combining the two cases, we conclude that $P \le \frac{3}{4}$, where equality holds if and only if $a = b = c = \frac{1}{2}$.
Here is a more brute-force approach using Lagrange multipliers. We seek to maximize the $C^1$ function $\def\RR{\mathbb{R}} f \colon \RR^3 \to \RR$ defined by $$ f(a, b, c) = ab + bc + ca $$ given the constraint $g(a, b, c) = 0$, where the $C^1$ function $g \colon \RR^3 \to \RR$ is defined by $$ g(a, b, c) = (a + b)(b + c)(c + a) - 1. $$ The gradients of $f$ and $g$ are $$ \nabla f(a, b, c) = \begin{bmatrix} b + c \\ c + a \\ a + b \end{bmatrix} $$ and $$ \nabla g(a, b, c) = \begin{bmatrix} (b + c)(2a + b + c) \\ (c + a)(2b + c + a) \\ (a + b)(2c + a + b) \end{bmatrix} \neq \mathbf{0}. $$ For an optimal solution $(a, b, c)$, there is a unique Lagrange multiplier $\lambda$ such that $\nabla f(a, b, c) = \lambda \nabla g(a, b, c)$, i.e., $$ \begin{bmatrix} b + c \\ c + a \\ a + b \end{bmatrix} = \lambda \begin{bmatrix} (b + c)(2a + b + c) \\ (c + a)(2b + c + a) \\ (a + b)(2c + a + b) \end{bmatrix}, $$ so $$ 2a + b + c = 2b + c + a = 2c + a + b = \frac{1}{\lambda} $$ (note that $\lambda > 0$), hence $a = b = c$. But $g(a, b, c) = 0$, so $(2a)^3 = 1$, hence $a = b = c = \frac{1}{2}$, and $\lambda = \frac{1}{2}$.
The bordered Hessian is $$ \begin{multline} \mathbf{H}(\lambda, a, b, c) \\ = \begin{bmatrix} 0 & (b + c)(2a + b + c) & (c + a)(2b + c + a) & (a + b)(2c + a + b) \\ (b + c)(2a + b + c) & - 2 \lambda (b + c) & 1 - 2 \lambda (a + b + c) & 1 - 2 \lambda (a + b + c) \\ (c + a)(2b + c + a) & 1 - 2 \lambda (a + b + c) & - 2 \lambda (c + a) & 1 - 2 \lambda (a + b + c) \\ (a + b)(2c + a + b) & 1 - 2 \lambda (a + b + c) & 1 - 2 \lambda (a + b + c) & - 2 \lambda (a + b) \\ \end{bmatrix}, \end{multline} $$ so $$ \mathbf{H}(\tfrac12, \tfrac12, \tfrac12, \tfrac12) = \begin{bmatrix} 0 & 2 & 2 & 2 \\ 2 & - 1 & -\tfrac12 & -\tfrac12 \\ 2 & -\tfrac12 & - 1 & -\tfrac12 \\ 2 & -\tfrac12 & -\tfrac12 & - 1 \\ \end{bmatrix}, $$ whose 3rd and 4th leading principal minors have determinants $$ \begin{vmatrix} 0 & 2 & 2 \\ 2 & - 1 & -\tfrac12 \\ 2 & -\tfrac12 & - 1 \\ \end{vmatrix} = 4 > 0 $$ and $$ \begin{vmatrix} 0 & 2 & 2 & 2 \\ 2 & - 1 & -\tfrac12 & -\tfrac12 \\ 2 & -\tfrac12 & - 1 & -\tfrac12 \\ 2 & -\tfrac12 & -\tfrac12 & - 1 \\ \end{vmatrix} = -3 < 0. $$ Therefore, $a = b = c = \frac{1}{2}$ is a local maximum point, and hence the optimal solution to the optimization problem.
We have $(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc$ which results in $$P = \frac{1 + abc}{a + b + c}.$$
Using AM-GM, we have $$(a + b)(b + c)(c + a) \ge 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8abc$$ and $$(a + b)(b + c)(c + a)\le \left(\frac{a + b + b + c + c + a}{3}\right)^3 = \frac{8}{27}(a + b + c)^3.$$
Thus, we have $abc \le 1/8$ and $a + b + c\ge 3/2$.
Thus, we have $$P = \frac{1 + abc}{a + b + c} \le \frac{1 + 1/8}{3/2} = 3/4.$$
Also, when $a = b = c = 1/2$, we have $(a + b)(b + c)(c + a) = 1$ and $P = 3/4$.
Thus, the maximum of $P$ is $3/4$.