# Choose the variables so that the weighted symmetric polynomial is minimal.

I've been struggling with the following problem for hours:

Consider the expression $p^2\frac{x}{y+z}+q^2\frac{y}{x+z}+r^2\frac{z}{x+y}$, where $p,q,r>0$ are parameters. Choose $x,y,z\ge0$ so that the value of the expression is a minimum.

Any ideas to begin with? :/

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Are you sure $x = y = z = 0$ is allowed? Hint: the order of $p, q, r$ i.e. which is higher and lower is important.. – Macavity Mar 9 '13 at 15:18
Ofc if more than one of them is zero, then the expression is undefined. However, from that statement, eg. x=0 is permitted. – Healpowah Mar 9 '13 at 15:21
Ok - it's implicit I suppose. Still the hint is valid. – Macavity Mar 9 '13 at 15:23
I somehow get the impression that the expression is a minimum if the 3 additives are equal. Is this right? – Healpowah Mar 9 '13 at 15:51
@Macavity see my answer here: math.stackexchange.com/questions/318982/… – Ivan Loh Mar 10 '13 at 4:28

## 1 Answer

Converting to a new set of variables, we have
$p^2 u + q^2 v + r^2 w$ to be minimised, where $\dfrac{1}{1+u} + \dfrac{1}{1+v}+ \dfrac{1}{1+w} = 2$

From the constraint, we have $w = \dfrac{1- uv}{2 uv + u + v}$, so we can write the unconstrained problem as:
$$\text{Minimise} \quad M = p^2 u + q^2 v + r^2 \dfrac{1- uv}{2 uv + u + v}$$

This should be possible to address easily by taking derivatives w.r.t. $u, v$ and setting them to zero. Using this approach, I get one solution which could be positive: $$u = \frac{-p + q + r}{2p} \quad \text{and} \quad v = \frac{p-q + r}{2q} \quad \text{and} \quad w = \frac {p + q - r}{2 r}$$

(Note these need not be always positive, unless $p, q, r$ can form a triangle. Also notable is that the weights are smaller, larger the square.)

With this, the minimum value of $M$ is $\quad \dfrac{(p+q+r)^2}{2} - (p^2 + q^2 + r^2)$ which suggests there should be a nice inequality way of getting here as well.

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Thanks. :) But what happens if p,q,r don't form a triangle? eg. $p=1000$, $q=2$, $r=1$. I find that with $x,y,z\ge 0$, the minimum is $4$, with $x=0$, $y=1/2$, $z=1$. There also seems to be a minimum when p is relatively big to q and r. How can you deal with that case? :/ – Healpowah Mar 9 '13 at 20:54
Well, as this is really a convex optimisation problem in $u, v, w$ with convex constraints, in case the global minimum is not in the feasible region, the optimum is on the boundary. In this case, if $p$ is too large, the corresponding weight then has to be on the boundary - viz. the weight is zero, taking $p$ out of the objective function and the rest follows as mentioned earlier in comments with a minimum of $2qr$. – Macavity Mar 10 '13 at 3:27