<s>[INST] Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, taking into account the time zone). Here's one question that I've been stuck with:
With $a, b, c$ $\in \mathbb{R}^{+}$ such that $ab + bc + ca + abc =2$, find the max of
$ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$
I'm just curious what kind(s) of techniques these junior high students can use, since as far as I remember when I was at that age, calculus and derivatives aren't taught until high school.
 [/INST] Here Laplace multipliers, derivatives and gradients do not exist. Only multiplying and checking if something simplifies.
In this case, symmetry for simplifying the problem into:
$$
\mathcal{M}: 3{x+1 \over x^2+2x+2}\\
st. x^3+3x^2=2
$$
This holds because every term on the function is a sum of independent terms. In here symmetry is not a false friend.
Which reduces to finding the positive roots of $x^3+3x^2-2=0$. We do not know to solve a cubic equation, so we only can guess to break the polynomial. Our first guess is $x=-1$. Broken:
$$
(x^3+x^2+2x^2+2x-2x-2)/(x+1)=x^2+2x-2 
$$
And the others two are given by, yes, we know how to solve that already:
$$
x=\frac 12 (-2\pm\sqrt{4+8})=-1\pm \sqrt{3}
$$
Hence the only positive solution is $x=\sqrt{3}-1=a=b=c$ and $M=3{\sqrt{3}\over 3+1-2\sqrt{3}+2\sqrt{3}-2+2}=3{\sqrt{3}\over 4}\sim 1.3$.
What if they are not equal?. Suppose $a=b=1,c=x$:
$$
\mathcal{M}: \frac45+{x+1 \over x^2+2x+2}\\
st. 3x=2
$$
then$x=2/3$ and $M=\frac45+{2/3+1 \over 4/9+4/3+2}=\frac45+{15 \over 34}\sim 1.2<1.3$
Yes, symmetry here helps, and the solution is a maximum.
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