Coefficient tricks: maximization

Weighted Mean : We define the weighted mean with paramter $p\in\mathbb{R}$ , with coefficients $\lambda_1+\ldots+\lambda_n \in\mathbb{R}^{+*}$ such that $\lambda_1+\ldots+\lambda_n = 1$ , as the following : $P_p(\{a_i\},\{\lambda_i\}) = \left\{\begin{array}{cl} \prod_{i=1}^n a_i^{\lambda_i}& \text{if } p = 0,\\ \left(\sum_{i=1}^n \lambda_i a_i^p\right)^{\frac{1}{p}} & \text{if } p \neq 0. \end{array}\right.$

Weighted Mean Inequality : If $p,q\in\mathbb{R}$ with $p<q$ , and if $\lambda_1+\ldots+\lambda_n \in\mathbb{R}^{+*}$ with $\lambda_1+\ldots+\lambda_n = 1$ , for all positive real numbers $a_1,\ldots,a_n$ , we have : $P_p(\{a_i\},\{\lambda_i\}) \leq P_q(\{a_i\},\{\lambda_i\})$

Equality holds if and only if $a_1=\ldots=a_n$ $--------------------------------$

We have : $\sqrt a + \sqrt b + \sqrt c = \sqrt a + \frac{1}{2} \sqrt{4b} + \frac{1}{3} \sqrt{9c}.$ We cannot use the weighted mean inequality with $a,4b$ and $9c$ since $\frac{1}{2}+\frac{1}{3}+1\neq 1$ . We solve this issue by dividing everything by $1+\frac{1}{2}+\frac{1}{3} = \frac{11}{6}$ as the following : $\sqrt a + \sqrt b + \sqrt c = \frac{11}{6} \cdot \left(\frac{6}{11} \sqrt a + \frac{3}{11} \sqrt{4b} + \frac{2}{11} \sqrt{9c}\right)$

Therefore, by QM-AM weighted inequality, we have : $\begin{aligned} \sqrt{a}+\sqrt{b}+\sqrt{c} &\leq \frac{11}{6} \cdot \sqrt{\frac{6}{11}\cdot a + \frac{3}{11} \cdot 4b + \frac{2}{11} \cdot 9c}\\[1mm] & = \frac{11}{6} \cdot \sqrt{\frac{6}{11} \cdot (a+2b+3c)}\\[1mm] & = \sqrt{\frac{11 \cdot 60}{6}}\\[1mm] & = \sqrt{110} \approx 10.488 \end{aligned}$

Equality holds if and only if $a = 4b = 9c$ .

By Titu's lemma ,

$\begin{aligned} a + \frac b{\frac 12} + \frac c{\frac 13} & \ge \frac {\left(\sqrt a + \sqrt b + \sqrt c\right)^2}{1 + \frac 12 + \frac 13} \\ \implies \sqrt a + \sqrt b + \sqrt c & \le \sqrt{\frac {11}6 \times 60} = \sqrt{110} \approx \boxed{10.5} \end{aligned}$

Equality occurs when $a= \dfrac {360}{11}$ , $b = \dfrac a4$ , and $c = \dfrac a9$ .

Let $x = \sqrt{a}, y = \sqrt{b} , z = \sqrt{c}$ , then we want to maximize

$f(x, y, z) = x + y + z$

subject to $x^2 + 2 y^2 + 3 z^2 = 60$

The constraint can be rewritten as $x^2 + (\sqrt{2} y )^2 + (\sqrt{3} z )^2 = 60$

Now define the vector $\mathbf{x} = ( x, \sqrt{2} y , \sqrt{3} z )$ , then $\mathbf{x \cdot x } = 60$

On the other hand, our function can be expressed as

$f(x, y, z) = x + y + z = x + \dfrac{1}{\sqrt{2}} (\sqrt{2} y) + \dfrac{1}{\sqrt{3}} (\sqrt{3} z ) = \mathbf{y \cdot x }$

where

$\mathbf{y} = (1, \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{3}})$

Now from the Cauchy-Schwartz inequality, we know that

$\mathbf{x \cdot y } \le | \mathbf{x} ||\mathbf{y} | = \sqrt{\mathbf{ x \cdot x}} \hspace{4pt} \sqrt{\mathbf{ y \cdot y}}$

Thus,

$f(x, y, z) \le \sqrt{ 60 } \sqrt{ 1 + \dfrac{1}{2} + \dfrac{1}{3} } = \sqrt{60 } \sqrt{ \dfrac{11}{6} } = \sqrt{110} \approx \boxed{10.488}$

Equality occurs when $x = 2 y = 3 z$ , and since $x^2 + 2 y^2 + 3 z^2 = 60$ then $x^2 + \dfrac{1}{2} x^2 + \dfrac{1}{3} x^2 = 60$ , from which

$x^2 = \dfrac{ 60 }{11/6} = \dfrac{360}{11}$ . Thus $a = x^2 = \dfrac{360}{11}, b = y^2 = \dfrac{90}{11}, c = z^2 = \dfrac{40}{11}$