Let $p$ be a positive real number. What is the minimum of $$g(a,b,c):=\left(\dfrac{a}{b+c}\right)^p+\left(\dfrac{b}{c+a}\right)^p+\left(\dfrac{c}{a+b}\right)^p$$ where $a,b,c$ are positive real numbers?

When $p=1$, Nesbitt's inequality shows that the minimum is $3/2$, achieved when $a=b=c$.

When $p=1/2$, this question shows that the minimum arbitrarily approches $2$, achieved when $a=1$ and $b=c\rightarrow\infty$. Note that the minimum is not achieved when $a=b=c$.

So it seems there is a regime of values of $p$ where the minimum is achieved when $a=b=c$, and at some point it transitions to the regime where the minimum is achieved when $a=1$ and $b=c\rightarrow\infty$?

We can normalize the expression using $a+b+c=1$ to be


Is Lagrange multipliers the right tool to solve this? Unfortunately I'm not familiar with it.

  • $\begingroup$ For $a=b=c$, you get a value of $3/2^p$ (giving the $3/2$ for $p=1$); for $a\searrow0$, $b=c$, you get as close as $2$ as you want. The value of $p$ for which the "switch" between the two lower bounds happens is $p^\ast=\log2 3 -1 \simeq 0.585$. wolframalpha.com/input/… $\endgroup$ – Clement C. Nov 2 '14 at 21:31
  • $\begingroup$ For $p>1$, you can use this inequality to see that $g_p(a,b,c)$ attains minimum at $a=b=c$. $\endgroup$ – Quang Hoang Nov 2 '14 at 22:32

Here is a complete but somewhat messy answer.

Define the function $$g(x,y,z) = \left( \frac{x}{1-x} \right)^p + \left( \frac{y}{1-y} \right)^p + \left( \frac{z}{1-z} \right)^p$$ on the triangle $\Delta$ defined by $x + y + z = 1, x,y,z \geq 0$. It is a continuous function $\Delta \to [0,+\infty]$.

Since $\Delta$ is a closed bounded set, $g$ must attain a minimum somewhere. This minimum may be attained on the boundary of the triangle, or at an interior point. The method of Lagrange multipliers shows that a necessary condition for a minimum to occur at an interior point is that $\nabla f$ and $\nabla g$ be parallel, where $f(x,y,z) = x + y + z - 1$. We have $$\nabla g = \left(\frac{p}{x^2}\left(\frac{1}{x} - 1\right)^{-p-1}, \frac{p}{y^2}\left(\frac{1}{y} - 1\right)^{-p-1}, \frac{p}{z^2}\left(\frac{1}{z} - 1\right)^{-p-1} \right)$$ and $\nabla f = (1,1,1)$. The function $$h(t) = t^2\left( \frac{1}{t} - 1\right)^{p + 1} = t^{1-p}(1-t)^{1 + p}$$ is strictly monotonic on the interval $[0,1]$ unless $p \in (-1,1)$, and even in this case calculus shows that the function $h$ never takes the same value more than twice. Yet if $g$ is to attain a minimum at an interior point $(x,y,z)$, we must have $h(x) = h(y) = h(z)$ (if $p \ne 0$).

Thus for $p \leq -1$ and $p \geq 1$, if the minimum of $g$ is attained at an interior point, then it must be one at which $x = y = z$. The case $p = 0$ is trivial, and for $p \in (-1,1) - \{ 0 \}$, if the minimum is attained at an interior point, then at least two of $x$, $y$ and $z$ must be equal.

Therefore in all cases the minimum must occur either for, say, $x = 0$ or $y = z$.

$g$ is infinite on the boundary of $\Delta$ if $p < 0$, thus the minimum must be attained at an interior point in this case. When $p > 0$, the expression to be minimized on the boundary is of the form $u + 1/u$, where $u = (b/c)^p$, so it is clearly minimal when $b = c$, so that the minimum on the boundary is $2$. Combined with the previous considerations, this shows that the minimum is always attained at a point where $b = c$ (even if $a = 0$).

Now assume that $b = c$, and let $t = a/b$. The problem is reduced to minimizing $j(t) = (t/2)^p + 2(t + 1)^{-p}$ for $t \geq 0$, with $t = 0$ corresponding to the boundary case, and $t = 1$ corresponding to the case of $a = b = c$. We already know that for $p \leq -1$, the minimum is $j(1) = 3/2^p$, and that for $p \geq 1$, it must be either $j(0) = 2$ or $j(1) = 3/2^p$, but the latter is smaller, so it must be the latter. For $p = 0$, the minimum is obviously $3$, and this leaves only $p \in (-1,1) - \{0\}$. Assume $p$ is chosen this way.

Now some algebra shows that $j'(t)$ always has the same sign as $k(t) = p(1 + t - 2t^{(1-p)/(1+p)})$. The possible locations of a minimum for $j(t)$ can be found by studying $k(t)$ instead. Note that $k(0) = p$, $k(1) = 0$, $k''(t) > 0$ and $k(t) \to +\infty$ as $p \to +\infty$. We divide the study into different cases.

Case 1. $- 1 < p < 0$. The conditions $k(0) < 0$, $k(1) = 0$ and $k''(t) > 0$ show that $k(t) \leq 0$ if and only if $t \leq 1$. Thus the minimum of $j(t)$ occurs at $t = 1$, which is the case $a = b = c$.

Case 2. $1/3 < p < 1$. In this case, $k'(1) > 0$. Combined with $k(0) > 0$, $k(1) = 0$ and $k''(t) > 0$, this shows that there is some $c \in (0,1)$ such that $k(t) \geq 0$ if and only if $c \leq t \leq 1$. Thus $j$ has a minimum either at $t = 0$ or $t = 1$. Comparing $j(0) = 2$ and $j(1) = 3/2^p$, we see that it is the former if $1/3 < p < (\log 3/2)/\log 2$, the latter if $(\log 3/2)/\log 2 < p < 1$, and either if $p = (\log 3/2)/\log 2$.

Case 3. $p = 1/3$. In this case $k'(1) = 0$. Combined with $k(1) = 0$ and $k''(t) > 0$, this shows that $k(t) \geq 0$ everywhere, hence $j(t)$ has its minimum at $t = 0$.

Case 4. $0 < p < 1/3$. In this case $k'(1) < 0$. Combined with $k(1) = 0$, $k''(t) > 0$ and $k(t) \to +\infty$ for $t \to +\infty$, we find that there is some number $c > 1$ such that $k(t) \leq 0$ if and only if $1 \leq t \leq c$. Thus $j(t)$ has local minima at $t = 0$ and $t = c$, and these are the only possibilities for its overall minimum.

We have $j(0) = 2$. We will prove that $j(c) > 2$. We have $j(t) = A(t)+B(t)$, where $A(t) = (t/2)^p$ and $B(t) = 2(t + 1)^{-p}$. Now $A(c)B(c) = 2^{1-p}(1 + 1/c)^{-p} > 2^{1-2p}$ since $c > 1$. It follows that $j(c) = A(c) + B(c) > 2^{3/2 - p} > 2^{7/6} > 2$.

In conclusion, the minimum is attained in all cases either for $b = c$ and $a = 0$ or for $a = b = c$.


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