This problem has two parts: $a)$ Let $k>0$, find the minimum of the function $f(x,y)=x+y$ over the set S=$\{(x,y) \in \mathbb R^2_{> 0}:xy=k\}$. $b)$ Prove that for every $(x,y) \in \mathbb R^2_{> 0}$ the inequality $\frac{x+y} {2}\geq \sqrt{xy}$ holds.$$$$ I want to find the minimum of $f$ restricted to the level set $g(x,y)=k$. The function $f$ and $g$ are differentiable and $\nabla g(x,y)=(y,x)\neq0$ for every $(x,y) \in \mathbb R^2_{> 0}$, then I can apply lagrange multipliers to find the critical points of f restricted to that region. By the equation $\nabla f(x,y)=\lambda\nabla g(xy)$ I get that $(1,1)=\lambda (y,x)$, which means $x=y$. By the conditions of the problem, it is required that the points of the form $(x,x)$ live in the level set. So, the only critical point is $(\sqrt{k},\sqrt{k})$. $$$$My question is, how can I prove that at $(\sqrt{k},\sqrt{k})$, $f_S$ attains a minimum?$$$$For part $b)$, let's assume I could prove that $f(\sqrt{k},\sqrt{k})$ is a minimum of $f_S$.Then, for all $(x,y) \in S$ it immediately follows that $x+y\ge \sqrt{k}+\sqrt{k}=2\sqrt{k}$, so dividing both sides by $2$, I get $\frac{x+y} {2}\ge\sqrt{k}=\sqrt{xy}$. I don't see why this inequality would also hold for $(x,y) \in \mathbb R^2_{>0}\setminus S$
3 Answers
For (b), suppose that $x,y\in\Bbb R_{>0}$, and let $k=xy$; from (a) you know that $x+y\ge 2\sqrt{k}$ and hence that $\frac{x+y}2\ge\sqrt{xy}$.
For (a) you don’t really need anything as fancy as Lagrange multipliers. You have $y=\frac{k}x$, so $x+y=x+\frac{k}x$; its derivative with respect to $x$ is $1\frac{k}{x^2}$, which is increasing on $x>0$ and equal to $0$ at $x=\sqrt{k}$, so you have a minimum.
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You're right, I was complicating it all. Thanks! – user100106 Oct 16 '13 at 05:05

@user100106: You’re welcome! – Brian M. Scott Oct 16 '13 at 05:06
For (a), it is easier to use $y=\frac{k}{x}$ and minimize $\phi(x)= x+\frac{k}{x}$ on $(0,\infty)$. Since $\lim_{x \downarrow 0} \phi(x) = \lim_{x \to \infty} \phi(x) = \infty$, we see that $\phi$ has a minimum in $(0,\infty)$, hence $\phi'(x) = 0$ at this point. This gives $x = \sqrt{k}$, from which we get $\phi(x) \ge 2 \sqrt{k}$ for all $ x > 0$.
In particular, $x+y \ge 2 \sqrt{k}$ for all $x,y >0$ satisfying $xy = k$. Hence $x+y \ge 2 \sqrt{xy}$.
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You already had proven that the point $ \ (\sqrt{k} \ , \sqrt{k}) \ $ is at the minimum when you made your gradient calculation. You found that this point lies on the "level surface" $ \ x \ + \ y \ = \ 2\sqrt{k} \ $ , which represents a plane through the point with gradient $ \ \nabla f \ = \ \langle 1 \ , \ 1 \rangle \ $ there (really, everywhere on the plane).
For a line with slope $ \ m \ $ in the $ \ xy$ plane, its direction vector is $ \ \langle 1 \ , \ m \rangle $ ; the associated unit vector is u = $ \ \frac{1}{\sqrt{1 + m^2}} \ \langle 1 \ , \ m \rangle \ $ . Consequently, the directional derivative at $ \ (\sqrt{k} \ , \sqrt{k}) \ $ is
$$ D_{\mathbf{u}} \ f \ \ = \ \ \nabla f \ \centerdot \ \mathbf{u} \ \ = \ \ \langle 1 \ , \ 1 \rangle \ \centerdot \ \frac{1}{\sqrt{1 + m^2}} \ \langle 1 \ , \ m \rangle \ \ = \ \ \frac{1 \ + \ m}{\sqrt{1 + m^2}} \ \ , $$
which is positive for $ \ 1 \ + \ m \ > \ 0 \ \ \Rightarrow \ \ m \ > \ 1 \ \ . $ Since $ \ S \ $ is restricted to the first quadrant, this condition is certainly met by any ray within this quadrant extending from $ \ (\sqrt{k} \ , \sqrt{k}) \ . $ So the directional derivative of $ \ f(x,y) \ $ in the level surface at this point is always positive, which tells us that the point marks the minimum for $ \ x \ + \ y \ $ on $ \ xy \ = \ k \ $ .
[Note that we are abandoned by a "Hessian matrix" calculation, since the plane $ \ x \ + \ y \ = \ 2\sqrt{k} \ $ has no critical points; the second partial derivatives tell us nothing of use.]
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