Maximization

Relevant wiki: Lagrange Multipliers

Target Function : $\displaystyle f(a,b,c) = ab+bc+ca$ subject to constraint $\displaystyle a+b+c=9$ .

We have $\displaystyle Lf(a,b,c,\lambda) = ab+bc+ca-\lambda(a+b+c-9)$

Evaluating the partial derivatives we get the four equations :

$\displaystyle \begin{cases} b+c=\lambda \\ c+a=\lambda \\ a+b=\lambda \\ a+b+c=9 \end{cases}$ , Solving we get $a=b=c=3$ .

If we end it here it is incomplete, we must determine what the critical points leads us to, a maxima,minima or saddle point.

Taking the Hessian Matrix of $Lf(a,b,c,\lambda)$ we see,

$\displaystyle\Delta = \begin{pmatrix} 0 & 1 & 1 & -1 \\ 1 &0 &1 &-1 \\ 1 & 1 & 0 & -1 \\ 1&1&1&0 \end{pmatrix}$ which is Negative definite matrix . Since $\displaystyle \Delta < 0$ so $f(a,b,c)$ attains it's maximum at $(a,b,c)=(3,3,3)$

So, $f_{\text{max}}(a,b,c) = 27$

Relevant wiki: Trivial Inequality

Firstly note that $a^{2}+b^{2}+c^{2}\ge ab+bc+ca$ .

Since $a+b+c=9 \implies a^{2}+b^{2}+c^{2}=81-2(ab+bc+ca)$ .

Thus $a^{2}+b^{2}+c^{2}=81-2(ab+bc+ca)\ge ab+bc+ca \implies 27 \ge ab+bc+ca$ .

Lemma: $a^2+b^2+c^2 ≥ ab+ac+bc \qquad (*)$

Proofs:

There are many proofs of this inequality,I shall provide three of them.

Proof 1: Consider the following inequalities: $\begin{aligned} a^2+b^2 &\stackrel{\text{AM-GM}}\geq 2ab\\ b^2+c^2&\stackrel{\text{AM-GM}}\geq 2bc\\ c^2+a^2&\stackrel{\text{AM-GM}}\geq 2ca \end{aligned}$ Adding all these,we get: $\begin{aligned} 2(a^2+b^2+c^2)≥ 2(ab+ac+bc)\\ \implies \boxed{a^2+b^2+c^2≥ab+ac+bc} \end{aligned}$

Proof 2:

Since $a,b,c$ are positive reals,and because the inequality is symmetric,we can assume WLOG that $a≥b≥c$ .Then we have,by the Rearrangment Inequality: $a\times a+b\times b+c\times c≥ a\times b+b\times c+c\times a\\ \implies \boxed{a^2+b^2+c^2≥ab+ac+bc}$

Proof 3: Multiplying $(*)$ by 2,we get: $2a^2+2b^2+2c^2≥ 2ab+2bc+2ac\\ (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)≥0\\ (a-b)^2+(b-c)^2+(c-a)^2≥0$ Which is true by the Trivial Inequality.

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Returning to our original problem, notice that: $(a+b+c)^2=9^2\\ \implies a^2+b^2+c^2=81-2(ab+ac+bc)$ Plugging this in $(*)$ ,we get: $a^2+b^2+c^2≥ ab+ac+bc\\ \implies 81-2(ab+ac+bc)≥ ab+ac+bc\\ \implies \boxed{ab+ac+bc≤27}$ Equality occurs as $a=b=c=3$

Relevant wiki: Cauchy-Schwarz Inequality

Using Cauchy-Schwarz inequality , we have:

$\begin{aligned} (ab+bc+ca)^2 & \le (a^2+b^2+c^2)(a^2+b^2+c^2) \\ \implies ab+bc+ca & \le a^2+b^2+c^2 \quad \quad \quad \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{Equality occurs when }a=b=c=3} \\ ab+bc+ca & \le (a+b+c)^2 - 2(ab+bc+ca) \\ 3(ab+bc+ca) & \le (a+b+c)^2 \\ ab+bc+ca & \le \frac{1}{3}(a+b+c)^2 = \frac{81}{3} = \boxed{27} \end{aligned}$