Definition 5.8.1. Convex set.
A set \(C\) in \(\mathbb R^{n}\) is called convex, if for any \(\mathbf a, \mathbf b\in C\) and any \(t\in \mathbb R, 0\le t \le 1\text{,}\) we have \((1-t)\mathbf a + t \mathbf b \in C\text{.}\)
Section 5.8 Convexity and Some Applications
Convexity plays an important role in many extremal problems and inequalities. We include this section here to illustrate how some of the continuity and compactness arguments are used in the context of convex functions.
Subsection 5.8.1 Convex sets vs Convex Functions
The notion of a convex set in \(\mathbb R^{n}\) is more general than that of a convex function.
Geometrically, the set \(\{(1-t)\mathbf a + t \mathbf b: 0\le t \le 1\}\) is the line segment in \(\mathbb R^{n}\) with \(\mathbf a, \mathbf b\) as its ends. A convex set needs not have any interior point.
Definition 5.8.2. Convex Function.
A real-valued function \(f\) defined on a convex set \(C\) is called convex, if for any \(\mathbf a, \mathbf b\in C\) and any \(t\in \mathbb R, 0\le t \le 1\text{,}\) we have
\begin{equation*}
f((1-t)\mathbf a + t \mathbf b)\le (1-t) f(\mathbf a) + t f(\mathbf b).
\end{equation*}
\(f\) is called strictly convex if we have the strict inequality
\begin{equation*}
f((1-t)\mathbf a + t \mathbf b)\lt (1-t) f(\mathbf a) + t f(\mathbf b) \text{ for any } 0\lt t \lt 1.
\end{equation*}
\(f\) is called concave if \(-f\) is convex. Equivalently, the defining inequality above is reversed for a concave function.
Geometrically, if we construct a line in \(\mathbb R^{n}\times \mathbb R\supset C\times \mathbb R\) through the points \((\mathbf a, f(\mathbf a)), (\mathbf b, f(\mathbf b))\text{,}\) then it has parametric equation
\begin{equation*}
\mathbf x=((1-t)\mathbf a + t \mathbf b, (1-t) f(\mathbf a) + t f(\mathbf b)),
\end{equation*}
so \((1-t) f(\mathbf a) + t f(\mathbf b)\) is the “height of the line above the point” \((1-t)\mathbf a + t \mathbf b\text{.}\) When \(f\) is convex, \(f((1-t)\mathbf a + t \mathbf b)\) stays below the height at \((1-t)\mathbf a + t \mathbf b\) of the above line segment for \(0\le t\le 1\text{.}\) Here is a Desmos page illustrating this geometric property.
\(f\) is a convex function iff the set \(\{(\mathbf x, y): \mathbf x\in C, y\ge f(\mathbf x)\}\) in \(\mathbb R^{n}\times \mathbb R\text{,}\) called the epigraph of \(f\text{,}\) is convex. Another characterization of a convex function is that for every real number \(c\text{,}\) the sub level set of \(f\) defined by \(\{\mathbf x: f(\mathbf x)\le c\}\) is a convex set.
Because of this relation, properties of convex functions can often be studied as properties of convex functions. We will later discuss briefly the notion of supporting hyperplane of a convex set and that of the graph of a convex function.
Subsection 5.8.2 Review of Properties of Univariate Convex Functions
The illustration above in the previous subsection also includes a sketch of the argument that the slope of the secant lines on a convex function of a single variable is an increasing function. Geometrically it seems clear that if \(f\) is a convex function of a single variable on an interval that contains \(a\lt b \lt c\text{,}\) then \(f(c)\) stays above the secant line through \((a, f(a), (b, f(b))\text{.}\) This can be derived from the above property of secant lines: for \(c>b\text{,}\)
\begin{equation*}
\frac{f(c)-f(a)}{c-a} \ge \frac{f(b)-f(a)}{b-a} \Leftrightarrow f(c)\ge f(a)+ \frac{f(b)-f(a)}{b-a}(c-a).
\end{equation*}
These inequalities also hold when \(c\lt a\text{.}\)
The continuity of a convex function of one variable at an interior point is proved using these bounds by linear functions. Say, \(a\) is an interior point. Then there exist \(b, c\) in the domain such that \(b>a>c\text{,}\) and the above property of secant lines implies that for any \(x, b>x>a\text{,}\)
\begin{equation*}
f(a)+ \frac{f(c)-f(a)}{c-a}(x-a)\le f(x)\le f(a)+ \frac{f(b)-f(a)}{b-a}(x-a).
\end{equation*}
Then the sandwich theorem implies that \(f(x)\to f(a)\) as \(x\to a+\text{.}\) The direction when \(x\to a-\) is done in a similar way.
Question 5.8.3. How can we extend this argument to higher dimensions?
A convex function is certainly continuous at an interior point when constrained along any one-dimensional lines, but there are infinitely many lines through any given point. Here we will see some ideas of compactness at play.
Solution.
We will assume that the origin is in the interior of the domain of \(f\text{;}\) in fact, we will assume the domain of \(f\) includes the unit ball centered at the origin, and describe ideas to prove the continuity of \(f\) at the origin.
On key idea is that there exist a finite number of points on the unit sphere, in fact \(2n\) points, such that any \(\bx\) with sufficiently small \(\Vert \bx\Vert\) can be written as
\begin{align}
\bx\amp =s\bar \bx \text{ for some
$s, 0 \le s \le 1$ and}\tag{5.8.1}\\
\bar\bx\amp = t_{1}\mathbf a_{1}+\cdots+ t_{n}\mathbf a_{n} \text{
for some $t_{i}\ge 0, t_{1}+\cdots+ t_{n}=1$}\text{,}\tag{5.8.2}
\end{align}
where \(\{\mathbf a_{1}, \ldots, \mathbf a_{n}\}\) are among the \(2n\) points on the unit sphere. Technically
\begin{equation*}
(x_{1},x_{2},\ldots, x_{n})=|x_{1}|\tilde \be_{1}+|x_{2}|\tilde\be_{2}+\ldots+|x_{n}|\tilde\be_{n},
\end{equation*}
where \(\tilde\be_{i}=\text{sgn} (x_{i}) \be_{i}\text{.}\) If we set \(s=|x_{1}|+|x_{2}|+\ldots+|x_{n}|\text{,}\) and when \((x_{1},x_{2},\ldots, x_{n})\ne (0, 0,\ldots, 0)\text{,}\) \(t_{i}=|x_{i}|/s\text{,}\) and \(\mathbf a_{i}= \tilde\be_{i}\text{,}\) \(\bar\bx=t_{1} \tilde\be_{1}+ t_{2}\tilde\be_{2}+\ldots+ t_{n}
\tilde\be_{n}\text{,}\) then we get (5.8.1), (5.8.2). The condition \(s\le 1\) is satisfied when \(\Vert \mathbf x \Vert \le \frac{1}{\sqrt n},\) for, by the Cauchy-Schwarz inequality
\begin{equation*}
s=|x_{1}|+\cdots+|x_{n}|\le \sqrt n \sqrt{ x_{1}^{2}+\cdots+x_{n}^{2}}\le 1.
\end{equation*}
It then follows that
\begin{align*}
f(\mathbf x) \le \amp [1-s] f(\mathbf 0)+ s f(\bar\bx)\\
\le \amp [1-s] f(\mathbf 0)+ s\left[ t_{1}f(\mathbf a_{1})+\cdots+
t_{n}f(\mathbf a_{n}) \right]\\
\le \amp f(\mathbf 0) +s\left[M-f(\mathbf 0)\right].
\end{align*}
where \(M\) is chosen so that \(f( \tilde \be_{i})\le M\) for all \(i\text{.}\) This implies that
\begin{equation*}
f(\mathbf x)-f(\mathbf 0)\le (M -f(\mathbf 0)) ( |x_{1}|+\cdots+|x_{n}|)\to 0 \text{ as }
\mathbf x \to 0.
\end{equation*}
We can certainly cover \(\mathbb S^{n-1}\) by a finite number of similarly constructed sets, and carry out this argument, which would allow us to prove that
\begin{equation*}
\limsup_{\mathbf x\to \mathbf 0 }f(\mathbf x)\le f(0).
\end{equation*}
To prove \(\liminf_{\mathbf x\to \mathbf 0}f(\mathbf x)\ge f(0)\text{,}\) we use the property of the secant lines as done in the one-dimensional case. For any \(\mathbf x\) such that \(\Vert \mathbf x \Vert \le \frac{1}{\sqrt n}\text{.}\) We bound \(f(\mathbf x)\) from below by the secant line through \((-\frac{\mathbf x}{\Vert \mathbf x \Vert \sqrt n}, f(-\frac{\mathbf x}{\Vert \mathbf x \Vert \sqrt n})),
(\mathbf 0, f(\mathbf 0))\text{:}\)
\begin{equation*}
f(\mathbf x)-f(\mathbf 0)
\ge \frac{f(\mathbf 0)-f(-\frac{\mathbf x}{\Vert \mathbf x \Vert \sqrt n})}{(\sqrt n)^{-1}}
\Vert \mathbf x \Vert.
\end{equation*}
Since the slope \(\frac{f(\mathbf 0)-f(-\frac{\mathbf x}{\Vert \mathbf x \Vert \sqrt n})}{(\sqrt n)^{-1}}\) has a lower bound due to the upper bound of \(f(-\frac{\mathbf x}{\Vert \mathbf x \Vert \sqrt n})\text{,}\) this allows us to conclude that \(\liminf_{\mathbf x\to \mathbf 0}f(\mathbf x)\ge f(0)\text{.}\)
The secant line property of a convex function implies that, if \([a, a+\epsilon]\) is in the domain of a convex function, then the slope of the secant line \(\frac{f(x)-f(a)}{x-a}\) has a limit as \(x\to a+\text{,}\) although this limit could be \(-\infty\text{.}\) If \(a\) is an interior point of the domain, then picking some \(c\lt a\) in the domain implies a lower bound of \(\frac{f(x)-f(a)}{x-a}\) in terms of \(\frac{f(c)-f(a)}{c-a}\) when \(x>a\text{,}\) so in such a case, \(f\) has a finite left derivative \(D_{-}f(a)\) and right derivative \(D_{+}f(a)\) at \(a\text{,}\) and \(D_{-}f(a)\le D_{+}f(a)\text{.}\) Furthermore, for any \(k, D_{-}f(a)\le k \le D_{+}f(a)\text{,}\)
\begin{align*}
\frac{f(x)-f(a)}{x-a} \ge D_{+}f(a) \ge k, \amp \text{ for } x> a; \\
\frac{f(x)-f(a)}{x-a} \le D_{-}f(a)\le k, \amp \text{ for } x\lt a.
\end{align*}
This then implies that
\begin{equation*}
f(x)\ge f(a)+k(x-a) \text{ for all } x \text{ in the domain of } f.
\end{equation*}
Since the right hand side, \(f(a)+k(x-a)\text{,}\) represents a straight line, the above inequality shows that a convex function has a (linear) support function at any interior point of its domain.
The support function property of a convex function can be used to give a simple proof of Jensen’s inequality.
Theorem 5.8.4. Jensen’s inequality.
Suppose that \(f:X\mapsto (A, B)\) and \(p(x)\ge 0\) is a density function on \(X\text{,}\) namely, \(\int_{X}p(x)\, dx=1\text{.}\) Suppose that \(\phi:(A, B) \mapsto \mathbb R\) is convex, then
\begin{equation}
\phi \left( \int_{X} f(x)p(x)\, dx \right)\le \int_{X}\phi(f(x))p(x)\, dx.\tag{5.8.3}
\end{equation}
In words, “\(\phi\) evaluated at the average of \(f\) is not more than the average of \(\phi\circ f\text{.}\)”
Proof.
Set \(\bar f= \int_{X} f(x)p(x)\, dx\text{.}\) It is easy to rule out the possibility that \(\bar f= A\) or \(B\text{,}\) so we may assume that \(A\lt \bar f\lt B\text{.}\) Using the support property of \(\phi\) at \(\bar f\text{,}\) there exists some \(k\) such that
\begin{equation*}
\phi (y) \ge \phi (\bar f) + k(y - \bar f) \text{ for all } y \in (A, B).
\end{equation*}
Substituting \(y \) by \(f(x)\text{,}\) multiplying the above inequality by \(p(x)\) and integrating over \(x\in X\text{,}\) we get
\begin{equation*}
\int_{X}\phi(f(x))p(x)\, dx\ge \phi (\bar f) \int_{X}p(x)\, dx + k\left( \int_{X} f(x)p(x)\, dx
- \bar f \int_{X}p(x)\, dx\right).
\end{equation*}
The right hand side is simply \(\phi (\bar f) \text{,}\) which proves the Jensen’s inequality.
Commonly used cases of Jensen’s inequalities include \(\phi(y)=-\ln y\) or \(y \ln y\text{.}\) Proofs for Hlder’s and Minkowski’s inequalities also use convexity in crucial ways.
Exercise 5.8.5. .
Prove that
\begin{equation*}
\ln \left( \int_{X} e^{u(x)}p(x)\, dx\right) \ge
\int_{X} u(x)p(x)\, dx
\end{equation*}
for \(p(x)\ge 0, \int_{X}p(x)\, dx=1\)
Subsection 5.8.3 Some Properties of Convex Functions of Several Variables
When proving the continuity of a convex function of several variables, we already saw the complications for multi-dimensions. We do not intend to do a serious study of properties of convex functions of several variables, but only want to briefly discuss a few properties related to the notion of supporting planes to illustrate how the notion of compactness comes into play.
Definition 5.8.6. Supporting Hyperplane of a Convex Set.
Let \(K\) be a convex set, \(\mathbf x_{0}\) be a point on the boundary of \(K\text{.}\) \(K\) is said to have a supporting hyperplane at \(\mathbf x_{0}\) if there exists a vector \(\mathbf n\) such that
\begin{equation*}
(\mathbf x - \mathbf x_{0})\cdot \mathbf n \ge 0 \text{ for all } \mathbf x \in K.
\end{equation*}
Definition 5.8.7. Supporting Hyperplane of a Convex Function.
Let \(K\) be a convex set and \(f(\mathbf x) \) be a convex function defined on \(K\text{.}\) Let \(\mathbf x_{0}\in K\text{.}\) The graph of \(f\) at \((\mathbf x_{0}, f(\mathbf x_{0}))\) is said to have a supporting hyperplane if there exists a non-zero vector \(\mathbf v\) such that
\begin{equation*}
f(\mathbf x) \ge f(\mathbf x_{0}) +\mathbf v \cdot (\mathbf x- \mathbf x_{0}) \text{ for all }
\mathbf x \in K.
\end{equation*}
Note that we use the same terminology in these two contexts, but they have a slight distinction, as illustrated by the simple example \(f(x)=-\sqrt{x}\) on \([0, 1]\text{.}\) As a function it does not have a supporting hyperplane (a straight line here) at \(x=0\text{,}\) but its epigraph has a supporting hyperplane at \((0, 0)\) (a vertical line).
For a convex function of a single variable we gave a proof of the existence of a supporting line at any interior point using the property of secant lines. We can apply this argument along any direction to a convex function of several variables, but it alone would not give us a supporting hyperplane at a point on the graph. The extension to multi-dimensions would necessarily involve some kind of limiting argument and compactness. We will discuss the following theorems.
Theorem 5.8.8. Existence of a Supporting Hyperplane of a Closed Convex Set.
Any boundary point of a closed convex set has a supporting hyperplane.
Proof.
Let \(\mathbf x_{0}\in K\) be a boundary point of the closed convex set \(K\text{.}\) Then there exists a sequence \(\mathbf x_{k} \not\in K, \mathbf x_{k} \to \mathbf x_{0}\text{.}\) Each \(\mathbf x_{k}\) also has a closest point \(\mathbf p_{k}\in K\text{.}\) This is done either by the Bolzano-Weierstrass compactness theorem or the parallelogram law of the Euclidean norm
\begin{equation*}
2\Vert \frac{\mathbf p -\mathbf q}{2}\Vert^{2}=\Vert \mathbf p- \mathbf x_{k}\Vert^{2}
+ \Vert \mathbf q- \mathbf x_{k}\Vert^{2}-2 \Vert \frac{\mathbf p +\mathbf q}{2}- \mathbf x_{k}\Vert^{2}
\end{equation*}
and the completeness of \(\mathbb R^{n}\text{.}\) This law shows that if \(\mathbf q_{l}\in K\) is such that
\begin{equation*}
\Vert \mathbf q_{l}- \mathbf x_{k} \Vert \to \inf \{\Vert \mathbf x- \mathbf x_{k} \Vert: \mathbf x \in K\},
\end{equation*}
then \(\mathbf q_{l}\) is a Cauchy sequence, therefore has a limit, and that limit is in \(K\text{.}\)
Next we claim that
\begin{equation}
(\mathbf p_{k}- \mathbf x_{k})\cdot (\mathbf x - \mathbf p_{k})\ge 0 \text{ for all } \mathbf x \in K.\tag{5.8.4}
\end{equation}
This follows from considering
\begin{equation*}
h(t) :=(t\mathbf x +(1-t) \mathbf p_{k}-\mathbf x_{k})\cdot (t\mathbf x +(1-t) \mathbf p_{k}-\mathbf x_{k}).
\end{equation*}
Note that \(h(t)=\Vert t\mathbf x +(1-t) \mathbf p_{k}-\mathbf x_{k}\Vert^{2}\text{,}\) and \(t\mathbf x +(1-t) \mathbf p_{k}\in K\) for \(0\le t \le 1\text{,}\) so \(h(0)\le h(t)\) for all \(0 \le t\le 1\text{.}\) It follows that
\begin{equation*}
h'(0)=2 (\mathbf p_{k}- \mathbf x_{k})\cdot (\mathbf x - \mathbf p_{k})\ge 0.
\end{equation*}
Define \(\mathbf n_{k}=(\mathbf p_{k}- \mathbf x_{k})/\Vert \mathbf p_{k}- \mathbf x_{k}\Vert\text{.}\) Then \(\mathbf n_{k}\) is a sequence of unit vectors, so there exists a subsequence, still denoted by itself, and a limiting unit vector \(\mathbf n\) such that \(\mathbf n_{k}\to \mathbf n\text{.}\) We also know that \(\mathbf p_{k} \to \mathbf x_{0}\) as
\begin{equation*}
\Vert \mathbf p_{k} - \mathbf x_{0} \Vert \le \Vert \mathbf p_{k} - \mathbf x_{k} \Vert
+ \Vert \mathbf x_{k} - \mathbf x_{0} \Vert \le 2 \Vert \mathbf x_{k}-\mathbf x_{0} \Vert.
\end{equation*}
For each fixed \(\mathbf x \in K\text{,}\) dividing through both sides of (5.8.4) by \(\Vert \mathbf p_{k}- \mathbf x_{k}\Vert\text{,}\) and passing to the limit, we get
\begin{equation*}
\mathbf n \cdot ( \mathbf x - \mathbf x_{0}) \ge 0,
\end{equation*}
which is the inequality defining a supporting hyperplane.
In summary, the idea is that, in the absence of a direct construction of a supporting plane at the given point, one finds a relatively easy way to construct a supporting plane at a nearby, but unspecified point, and one then takes a limiting process to obtain a supporting plane at the given point.
Theorem 5.8.9. Existence of a Supporting Hyperplane of the Graph of a Convex Function at an Interior Point.
Let \(f(\mathbf x) \) be a convex function defined on the convex set \(K\text{.}\) If \(\mathbf x_{0}\in K\) in an interior point of \(K\text{,}\) then the graph of \(f\) has a supporting hyperplane at \((\mathbf x_{0}, f(\mathbf x_{0}))\text{.}\)
Proof.
The epigraph \(G_{f}=\{(\mathbf x, y):\mathbf x\in K, y\ge f(\mathbf x)\}\) is a convex set. Its closure \(\overline{G_{f}}\) is a closed convex set, and \((\mathbf x_{0}, f(\mathbf x_{0}))\) is on the boundary of \(\overline{G_{f}}\text{.}\) By the previous theorem, there exists a non-zero vector \(\mathbf n=(\mathbf v, c)\) such that
\begin{equation*}
\mathbf v \cdot (\mathbf x- \mathbf x_{0}) + c (y-f(\mathbf x_{0}))\ge 0 \text{ for all }
(\mathbf x, y) \in \overline{G_{f}}.
\end{equation*}
Since \(\mathbf x_{0}\in K\) in an interior point of \(K\text{,}\) we claim that \(c\ne 0\text{.}\) For, otherwise, we would have
\begin{equation*}
\mathbf v \cdot (\mathbf x- \mathbf x_{0})\ge 0 \text{ for all } \mathbf x\in K,
\end{equation*}
which would force \(\mathbf v=\mathbf 0\text{.}\)
Next we claim that \(c>0\text{.}\) This is because \((\mathbf x_{0}, f(\mathbf x_{0})+t)
\in G_{f}\) for any \(t>0\text{,}\) and the above inequality then forces \(c>0\text{.}\) Now it follows that for any \(\mathbf x\in K\text{,}\) applying the above inequality for \(y=f(\mathbf x)\) implies that
\begin{equation*}
f(\mathbf x) \ge f(\mathbf x_{0}) -c^{-1} \mathbf v \cdot (\mathbf x- \mathbf x_{0}),
\end{equation*}
which demonstrates a supporting hyperplane to the graph of \(f\) at \((\mathbf x_{0}, f(\mathbf x_{0}))\text{.}\)
It is possible to prove this theorem directly using the properties of a convex function, along the lines of proof for the one dimensional case. You should try to construct such a proof, at least for the two dimensional case.
Exercise 5.8.10.
The tangent plane of a convex function at a differentiable point is a supporting plane to the graph of the function. Furthermore, if the point is in the interior of the domain, then it is the unique supporting plane
Hint.
If \(f\) denotes the function, and \(Df(\mathbf x_{0})\) denotes the gradient of \(f\) at \(\mathbf x_{0}\text{,}\) it may be geometrically easier to consider
\begin{equation*}
g(\mathbf x)=f(\mathbf x)-f(\mathbf x_{0})-Df(\mathbf x_{0})\cdot (\mathbf x-\mathbf x_{0})\text{,}
\end{equation*}
which is also convex.
We close this subsection by discussing a more subtle application of convex/concave functions in an optimization problem.
Example 5.8.11.
Using concavity to identify the minimum of \(\frac{ \Big| \sum_{i=1}^{n}a_{i}b_{i}\Big|}{\left(\sum_{i=1}^{n}a_{i}^{2}\right)^{1/2}
\left(\sum_{i=1}^{n}b_{i}^{2}\right)^{1/2}}\) under some constraints on \((a_{1},\ldots, a_{n}), (b_{1},\ldots, b_{n})\)
We introduce the new variables \(u_{i}=a_{i}^{2}, v_{i}=b_{i}^{2}\text{,}\) and reformulate the problem in terms of \(u_{i}, v_{i}\text{.}\) The quotient then becomes
\begin{equation*}
\frac{\sum_{i} \sqrt{u_{i} v_{i}}}{\sqrt{(\sum_{i} u_{i}) (\sum_{i} v_{i})}},
\end{equation*}
and the constraints become
\begin{equation*}
a^{2} \le u_{i}\le A^{2}; b^{2}\le v_{i}\le A^{2}.
\end{equation*}
Our argument will be based on the following observation.
For the second item, note that for any \((u, v)\) in the rectangle \([a^2, A^2]\times[b^2, B^2]\text{,}\) there exists a unique \(s>0\text{,}\) such that \(s (u, v)\) lies on the diagonal from \((a^2, B^2)\) to \((A^2, b^2)\text{,}\) which implies the existence of a unique \(0\le t \le 1\) such that
\begin{equation*}
s (u, v)= (1-t) (a^2, B^2) + t(A^2, b^2).
\end{equation*}
This then implies our desired relation.
We remark that in proving the last item above, only the (strict) concavity of \(\sqrt{uv}\) along the diagonal from \((a^2, B^2)\) to \((A^2, b^2)\) is used. It is this last item that makes it possible to bound \(\sum_{i} \sqrt{u_{i} v_{i}}\) from below.
Now for each \((u_{i}, v_{i})\text{,}\) we find \((p_{i}, q_{i})\) according to the second item above
\begin{equation*}
(u_{i}, v_{i})= p_{i} (a^2, B^2) + q_{i}(A^2, b^2),
\end{equation*}
then we can bound the quotient as
\begin{equation*}
\frac{\sum_{i} \sqrt{u_{i} v_{i}}}{\sqrt{(\sum_{i} u_{i}) (\sum_{i} u_{i})}}\ge
\frac{\sum_{i} (p_{i} aB+ q_{i} Ab)}
{\sqrt{[\sum_{i} (p_{i} a^2 +q_{i} A^{2})][ \sum_{i} (p_{i}B^{2}+q_{i}b^{2})]}}.
\end{equation*}
Setting \(p=\sum_{i} p_{i}, q=\sum_{i} q_{i}\text{,}\) \(\alpha=A/a, \beta=B/b\text{,}\) and after dividing both the numerator and denominator of the quotient on the right hand above by \(ab\text{,}\) it becomes
\begin{equation*}
\frac{p \beta + q \alpha}{\sqrt{ (p +q \alpha^{2})(p\beta^{2}+q)}},
\end{equation*}
and now the task is to find the infimum of this quotient when \(p, q\ge 0\text{,}\) and identify when equality can occur. This calculus problem can be solved in a routine way, and the answer is \(\frac{2\sqrt{\alpha \beta}}{ \alpha \beta+1}\text{.}\)
