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Consider $f: (\mathbb R_+)^n \to \mathbb R$ defined by

$$f(x_1,\ldots,x_n) = (1+x_1)\cdots(1+x_n)$$

I am looking for local or global extrema under the condition $x_1\cdots x_n=a^n, a >0$ using the method of Lagrange multipliers.

First of all I applied $\ln$ and I ended up with the function

$$H(x_1,\ldots,x_n,\lambda)=\sum_{i=1}^{n}\ln(1+x_i)+\lambda(\sum_{i=1}^{n}\ln(x_i)-n\ln(a))$$

So we get

$$\frac{\partial H}{\partial x_i}=\frac{1}{1+x_i}+\lambda\frac{1}{x_i}$$ and $$\frac{\partial H}{\partial \lambda}=\sum_{i=1}^{n}\ln(x_i)-n\ln(a)$$

Setting the 2nd equation equal to zero, we receive $$x_i=a,\forall i$$ and plugging in $x_i=a$ in the first equation delivers $$\lambda=-\frac{a}{1+a}$$

Is that correct so far?

How can I argue that this is indeed a extremum? I would like to use Cauchy-Schwarz inequality if possible.

Community
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user699386
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1 Answers1

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Yes, you are right!

For, example, it's enough to prove that $$\sum_{i=1}^n\left(\ln(1+x_i)-\ln(1+a)-\frac{a}{1+a}(\ln{x_i}-\ln{a})\right)\geq0,$$ for which it's enough to prove that: $$\ln(1+x_i)-\ln(1+a)-\frac{a}{1+a}(\ln{x_i}-\ln{a})\geq0$$ or $$\left(\frac{1+x_i}{1+a}\right)^{1+a}\geq\left(\frac{x_i}{a}\right)^a,$$ which is true by AM-GM: $$\left(\frac{1+x_i}{1+a}\right)^{1+a}=\left(\frac{1+a\cdot\frac{x_i}{a}}{1+a}\right)^{1+a}\geq\left(\frac{(1+a)\left(1\cdot\left(\frac{x_i}{a}\right)^a\right)^{\frac{1}{a+1}}}{1+a}\right)^{1+a}=\left(\frac{x_i}{a}\right)^a.$$

Michael Rozenberg
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  • So when checking if the possible extremum is indeed an extremum we can do this by working with the function $H$? I found that a bit confusing because the function we wanted to maximize or minimize is $f$ instead of $H$ – user700332 Oct 04 '19 at 05:36
  • @user700332 If we'll find a minimum of $H$, so we'll find a minimum of $f$. They are equivalent problems. – Michael Rozenberg Oct 04 '19 at 06:47