about Lagrange multipliers

Question

For the real numbers $a, b, c$ and $d$ such that: $a+b+c+d=4$ and $a^2+b^2+c^2+d^2=s \ge \frac {28}{3}$, I have to find the maximum value of the product $abcd$ in terms of $s.$ We may use the Lagrange multipliers, that means if $f(a,b,c,d)=abcd, g(a,b,c,d)=a+b+c+d-4$ and $h(a,b,c,d)=a^2+b^2+c^2+d^2-s,$ then

$bcd+λ_1+λ_2 a=0,$

$acd+λ_1+λ_2 b=0,$

$abd+λ_1+λ_2 c=0,$

$abc+λ_1+λ_2 d=0,$

or

$ \frac {f}{a}+λ_1+λ_2 a=0,$

$\frac {f}{b}+λ_1+λ_2 b=0,$

$\frac {f}{c}+λ_1+λ_2 c=0,$

$\frac {f}{d}+λ_1+λ_2 d=0.$

By adding them, we have: $f=-(λ_1+\frac{s λ_2}{4} )$ and I can't go further. On the other hand, I have a second thought:

$\sqrt {\frac{a^2+b^2+c^2+d^2}{4}} \ge \sqrt[4]{|a| |b| |c| |d|}=\sqrt[4]{|a b c d|}$

or

$\frac{s^2}{16} \ge |a b c d| \ge abcd$

Am I right?

Answer 1

You can end it after writing $$abcd+\lambda_1a+2\lambda_2a^2=abcd+\lambda_1b+2\lambda_2b^2=abcd+\lambda_1c+2\lambda_2c^2=abcd+\lambda_1d+2\lambda_2d^2=0,$$ which gives $$\lambda_1a+2\lambda_2a^2=\lambda_1b+2\lambda_2b^2=\lambda_1c+2\lambda_2c^2=\lambda_1d+2\lambda_2d^2$$ and from here $$(a-b)(\lambda_1+2\lambda_2(a+b))=(a-c)(\lambda_1+2\lambda_2(a+c))=(a-d)(\lambda_1+2\lambda_2(a+d))=$$ $$=(b-c)(\lambda_1+2\lambda_2(b+c))=(b-d)(\lambda_1+2\lambda_2(b+d))=(c-d)(\lambda_1+2\lambda_2(c+d)).$$ Now, easy to get all critical points and the answer $\frac{(s-8)^2}{16}.$

I like the following way.

We can assume that $abcd>0$, which gives two cases.

1. $a>0$, $b>0$, $c<0$ and $d<0$. We rewrite $c$ as $-c$ and $d$ as $-d$.

Thus, $c>0$ and $d>0$.

Now, let $a+b=4+c+d=k$.

Thus, by C-S $$s\geq\frac{1}{2}(a+b)^2+\frac{1}{2}(c+d)^2=\frac{1}{2}(k^2+(k-4)^2)=k^2-4k+8.$$ Also, by AM-GM $$abcd\leq\left(\frac{a+b}{2}\right)^2\left(\frac{c+d}{2}\right)^2=\frac{k^2}{4}\cdot\frac{(k-4)^2}{4}=\frac{(k^2-4k)^2}{16}\leq\frac{(s-8)^2}{16}.$$ The equality occurs for $a=b$, $c=d$ and $k^2-4k+8=s,$ which is possible for $s\geq\frac{28}{3}.$

The second case, when all variables are positives, for you.