# Find the values of $x$, $y$ and $z$ minimizing $\frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y}$

$$\frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y}$$

$r$, $s$, $t$ are positive coefficients. Find the values of non-negative variables $x$, $y$ and $z$ so that the above expression is a minimum.

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Solution 1: (without calculus)

Consider 2 cases:

Case 1: $r, s, t$ satisfy the triangle inequality. Then \begin{align} \frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y} & =-(r^2+s^2+t^2)+\frac{r^2(x+y+z)}{y+z}+\frac{s^2(x+y+z)}{x+z}+\frac{t^2(x+y+z)}{x+y} \\ & =-(r^2+s^2+t^2)+(x+y+z)(r+s+t)\left(\frac{r(\frac{r}{y+z})+s(\frac{s}{x+z})+t(\frac{t}{x+y})}{r+s+t}\right) \\ & \geq -(r^2+s^2+t^2)+(x+y+z)(r+s+t)\left(\frac{r+s+t}{r(\frac{y+z}{r})+s(\frac{x+z}{s})+t(\frac{x+y}{t})}\right) \\ & =-(r^2+s^2+t^2)+\frac{(r+s+t)^2}{2} \\ & =(rs+st+tr)-\frac{r^2+s^2+t^2}{2} \end{align}

where the inequality is by AM$\geq$HM. Equality holds when $\frac{r}{y+z}=\frac{s}{x+z}=\frac{t}{x+y}$, or equivalently, when $x=(s+t-r)a, y=(r+t-s)a, z=(r+s-t)a$, where $a \in \mathbb{R}^+$.

Edit: Point of Clarification: The above inequality holds even if $r, s, t$ do not satisfy the triangle inequality, but equality cannot be attained, since at least one of $s+t-r, r+t-s, r+s-t$ will be negative. As such, the inequality is strict, and we have not found the minimum.

Case 2: $r, s, t$ do not satisfy the triangle inequality. We shall WLOG assume $r+s \leq t$.

With the same argument as above, we have
\begin{align} &\frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y} \\ & \geq \frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{(r+s)^2z}{x+y} \\ & = -(r^2+s^2+(r+s)^2)+\frac{r^2(x+y+z)}{y+z}+\frac{s^2(x+y+z)}{x+z}+\frac{(r+s)^2(x+y+z)}{x+y} \\ & \geq -(r^2+s^2+(r+s)^2)+\frac{(r+s+(r+s))^2}{2} \\ &=2rs \end{align} with equality when $x=2sa, y=2ra, z=0, a \in \mathbb{R}^+$.

We get similar results for $r+t \leq s$ and $s+t \leq r$, with the variables permuted.

Solution 2: (using Lagrange multipliers)

As before, $$\frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y} =-(r^2+s^2+t^2)+\frac{r^2(x+y+z)}{y+z}+\frac{s^2(x+y+z)}{x+z}+\frac{t^2(x+y+z)}{x+y}$$

The inequality is homogeneous, so WLOG assume $x+y+z=1$, and the expression becomes $$-(r^2+s^2+t^2)+\frac{r^2}{1-x}+\frac{s^2}{1-y}+\frac{t^2}{1-z}$$

We first check end points. We have either at least 1 of them $=0$ or at least 1 of them $=1$, but since the latter case implies the former case, it suffices to check when at least 1 of them is $=0$. If $z=0$, the inequality becomes $$-(r^2+s^2+t^2)+\frac{r^2}{1-x}+\frac{s^2}{1-y}+t^2=\frac{r^2(1-y)}{1-x}+\frac{s^2(1-x)}{1-y} \geq 2rs$$ by AM$\geq$GM inequality. Similarly, when $x=0, y=0$, we get a minimum of $2st, 2rt$ respectively.

To get critical points, partial differentiate $-(r^2+s^2+t^2)+\frac{r^2}{1-x}+\frac{s^2}{1-y}+\frac{t^2}{1-z}+\lambda (x+y+z-1)$ w.r.t. $x, y, z, \lambda$, to get $\frac{r^2}{(1-x)^2}=\frac{s^2}{(1-y)^2}=\frac{t^2}{(1-z)^2}$, or equivalently $x=(s+t-r)a, y=(r+t-s)a, z=(r+s-t)a$, where $\frac{1}{a}=(s+t-r)+(r+t-s)+(r+s-t)$, which is possible when $r, s, t$ satisfy the triangle inequality. Checking, this gives $(rs+st+tr)-\frac{r^2+s^2+t^2}{2} \leq \min(2rs, 2rt, 2st)$, so the minimum is achieved at $x=(s+t-r)a, y=(r+t-s)a, z=(r+s-t)a$, where $a \in \mathbb{R}^+$, for the case where $r, s, t$ satisfy the triangle inequality.

On the other hand, if $r, s, t$ do not satisfy the triangle inequality, then we have no critical points, so $\frac{r^2x}{y+z}+\frac{s^2y}{x+z}+\frac{t^2z}{x+y}$ is minimised at $\min(2rs, 2rt, 2st)$. If $t=max(r, s, t)$, then we get a minimum of $2rs$, achieved when $z=0, x=\frac{s}{r+s}, y=\frac{r}{r+s}$. Removing the assumption $x+y+z=1$ gives the general form $x=sa, y=ra, a \in \mathbb{R}^+, z=0$ for the equality case.

We get similar results (with variables permuted) if $max(r, s, t)$ is $r$ or $s$.

Conclusion: I still prefer the approach without calculus. It is so much more elegant.

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Thanks for your help. The only thing is that I still don't get the triangle inequality part. Why is it necessary to separate these cases? I guess you're talking about weighted AM-HM inequality which, as far as I know, only needs the sum of the nonnegative weights $r$, $s$, $t$ to be greater than 0. – Anonymous Mar 3 '13 at 9:54
Yes, the inequality still holds, but the equality case cannot be attained when $r, s, t$ don't satisfy the triangle inequality, so that isn't the minimum. – Ivan Loh Mar 3 '13 at 9:55
Then I suppose $a$ has to be positive as well. – Anonymous Mar 3 '13 at 10:57
Ah yes, forgot to add that restriction. I'll edit that in. – Ivan Loh Mar 3 '13 at 11:15

Take partial derivatives in x, y, and z set each resulting derivative to zero, use the three equations that result to solve for x, y and z, those will be all of your critical points. Then use the hessian matrix to determine where minimum occurs.

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