<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] If $a=b=c=\sqrt3-1$ then we get a value $\frac{3\sqrt3}{4}$.
We'll prove that it's a maximal value.
Indeed, let $x=\frac{a+1}{\sqrt3}$, $y=\frac{b+1}{\sqrt3}$ and $z=\frac{c+1}{\sqrt3}$.
Hence, the condition gives $x+y+z=3xyz$ and we need to prove that
$$\sum_{cyc}\frac{x}{3x^2+1}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x}{3x^2+\frac{3xyz}{x+y+z}}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\frac{x+y+z}{3(x+y)(x+z)}\leq\frac{3}{4}$$ or
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)^2$$ and since
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ it's just 
$$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that
$$xy+xz+yz\geq x+y+z$$ or
$$xy+xz+yz\geq(x+y+z)\sqrt{\frac{3xyz}{x+y+z}}$$ or
$$(xy+xz+yz)^2\geq3xyz(x+y+z)$$ or
$$\sum_{cyc}z^2(x-y)^2\geq0.$$
Done!
</s>