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

Question

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.

Answer 1

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!

Answer 2

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.

Answer 3

The functional form of $M$ suggests us to change variable to $u = a+1, v = b+1, w = c + 1$.
In terms of them, the expression of $M$ simplifies to

$$M = \frac{u}{1+u^2} + \frac{v}{1+v^2} + \frac{w}{1+w^2}$$

The constraint $abc + ab + bc + ca = 2$ can be rewritten as

$$(a+1)(b+1)(c+1) - (a+b+c) - 1 = 2\quad\iff\quad u + v + w - uvw = 0$$

Since $a,b,c > 0$, $u,v,w \in (1,\infty)$. Pick $A, B, C \in (\frac{\pi}{4},\frac{\pi}{2})$ such that $u = \tan A$,$v = \tan B$ and $w = \tan C$. The constraint tell us $$\tan(A+B+C) = \frac{u+v+w - uvw}{1 - uv - vw - wu} = 0 \quad\implies\quad A+B+C = k\pi, \quad k \in \mathbb{Z} $$ Since $A + B + C \in (\frac{3\pi}{4}, \frac{3\pi}{2})$, $k$ equals to $1$ and $A + B + C = \pi$.

In terms of $A,B,C$, the expression we want to maximize becomes

$$M = \frac{\tan A}{1+\tan^2 A} + \frac{\tan B}{1+\tan^2 B} + \frac{\tan C}{1+\tan^2 C} = \frac12 \left[ \sin(2A) + \sin(2B) + \sin(2C) \right]$$

This is nothing but the area of triangle inscribed inside unit circle with angles $2A$, $2B$ and $2C$ subtended at center. It is well known for such triangles, the one which maximize the area is an equilateral triangle. This implies $M$ is maximized when

$$A = B = C = \frac{\pi}{3} \implies a = b = c = \tan\frac{\pi}{3} - 1 = \sqrt{3} - 1$$

and the maximum value of $M$ is $\displaystyle\;\frac{3}{2}\sin\frac{2\pi}{3} = \frac{3\sqrt{3}}{4}$.