Means and Lagrange multipliers

[gugo82]

The following is a very elementary exercise, an application of Lagrange method to prove a widely known set of inequalities.

***

Three days ago I had to lecture about how to solve constrained optimization problems and Lagrange multipliers method.

I was in trouble: in fact the textbook I usually refer to "draw inspiration" for exercises lacked of interesting/funny material... I was thinking about what to do when I suddenly remembered a set of nice inequalities between means.
I proved those inequalities sooo much time ago using simpler argument (namely, a standard technique of Calculus); but at that moment I realized that they can be also regarded as easy contrained optimization problems.

Then I sketched some passages and I made it: I got a nice set of meaningful exercises.*

***

Exercise:

1. Let [tex]$x,y \geq 0$[/tex] and put:

[tex]$A(x,y):=\frac{1}{2}(x+y),\ G(x,y):=\sqrt{x,y},\ H(x,y):=\frac{2}{\frac{1}{x} +\frac{1}{y}}$[/tex]

so that [tex]$A(x,y),\ G(x,y),\ H(x,y)$[/tex] are the arithmetic, geometric and harmonic mean of [tex]$x$[/tex] and [tex]$y$[/tex].**

Use Lagrange multipliers method to show that:

(AMGM) [tex]$A(x,y)\geq G(x,y)$[/tex],
(AMHM) [tex]$A(x,y)\geq H(x,y)$[/tex],
(GMHM) [tex]$G(x,y)\geq H(x,y)$[/tex],

with equality iff [tex]$x=y$[/tex] in (AMGM) and (AMHM) anf iff [tex]$x=y \text{ or } x=0 \text{ or } y=0$[/tex] in (GMHM).

2. Let [tex]$x,y\geq 0$[/tex], choose [tex]$p\in \mathbb{R} \setminus \{ 0\}$[/tex] and set:

[tex]$M_p(x,y):=\left\{ \frac{1}{2}(x^p+y^p)\right\}^\frac{1}{p}$[/tex];

we say that [tex]$M_p(x,y)$[/tex] is the [tex]$p$[/tex]-mean of [tex]$x$[/tex] and [tex]$y$[/tex].***

Using Lagrange multipliers method, show that for [tex]$p>q$[/tex] the following general inequality holds:

(pMqM) [tex]$M_p(x,y)\geq M_q(x,y)$[/tex]

and that equality case occurs in (pMqM) iff [tex]$x=y$[/tex] (in the case [tex]$p>0$[/tex]) or iff [tex]$x=y \text{ or } x=0 \text{ or } y=0$[/tex] (for [tex]$p<0$[/tex]).

3. Find an elementary Calculus proof of (pMqM).

4. Prove that [tex]$\max \{ x,y\} \geq M_p(x,y) \geq \min \{ x, y\}$[/tex] and:

(a) [tex]$\lim_{p\to +\infty} M_p(x,y) =\max \{ x, y\}$[/tex],
(b) [tex]$\lim_{p\to -\infty} M_p(x,y) =\min \{ x, y\}$[/tex],
(c) [tex]$\lim_{p\to 0} M_p(x,y)=G(x,y)$[/tex],

then realize that (pMqM) holds for all [tex]$p> q \in [-\infty ,+\infty]$[/tex] with the equality case characterized as in 2.

Hints: 1 & 2. Fixed [tex]$g\geq 0$[/tex], to prove (AMGM) it suffices to show that [tex]$A(x,y)\geq g$[/tex] on the constraint [tex]$G(x,y)=g$[/tex] (why?); use Lagrange method. This technique also works for (AMHM), (GMHM) and (pMqM).

3. It could be useful to observe that both righthand and lefthand sides of (pMqM) are (positively) homogeneous functions of degree [tex]$1$[/tex], therefore they depends only on the value of the ratio [tex]$t=\frac{y}{x}$[/tex].

4. The inequalities are trivial. To prove (a) and (b) it suffices to find the higher order infinite/lower order infinitesimal in [tex]$M_p(x,y)$[/tex] as [tex]$p\to \pm \infty$[/tex] and play with them. The limit in (c) comes in indeterminate form [tex]$1^\infty$[/tex]: try to solve it with the usual trick (i.e. [tex]$M_p(x,y)=e^{\ln M_p(x,y)}$[/tex]) and with an application of de l'Hôpital's rule.

__________
* Of course they're meaningful for me 'cause I learnt something thinking about them; perhaps they're not so for my "wannabe engineer"...

** [tex]H(x,y)[/tex] should be defined only for [tex]$x,y>0$[/tex]. But we have [tex]$\lim_{x\to 0^+} H(x,\overline{y})=\lim_{y\to 0^+} H(\overline{x},y)=\lim_{(x,y)\to (0,0)} H(x,y)=0$[/tex] (for all fixed [tex]$\overline{x},\overline{y}>0$[/tex]), therefore [tex]H(x,y)[/tex] can be extended by continuity on the whole set [tex]$x,y\geq 0$[/tex].

*** If [tex]$p<0$[/tex] then [tex]$M_p(x,y)$[/tex] is defined for [tex]$x,y>0$[/tex], but it can be exetended on the whole set [tex]$x,y\geq 0$[/tex] by continuity (as in **).

[gugo82]

"gugo82":4. Prove that [tex]$\max \{ x,y\} \geq M_p(x,y) \geq \min \{ x, y\}$[/tex] and:

(a) [tex]$\lim_{p\to +\infty} M_p(x,y) =\max \{ x, y\}$[/tex],
(b) [tex]$\lim_{p\to -\infty} M_p(x,y) =\min \{ x, y\}$[/tex],
(c) [tex]$\lim_{p\to 0} M_p(x,y)=G(x,y)$[/tex],

then realize that (pMqM) holds for all [tex]$p> q \in [-\infty ,+\infty]$[/tex] with the equality case characterized as in 2.

Hints: [...] 4. The inequalities are trivial. To prove (a) and (b) it suffices to find the higher order infinite/lower order infinitesimal in [tex]$M_p(x,y)$[/tex] as [tex]$p\to \pm \infty$[/tex] and play with them. The limit in (c) comes in indeterminate form [tex]$1^\infty$[/tex]: try to solve it with the usual trick (i.e. [tex]$M_p(x,y)=e^{\ln M_p(x,y)}$[/tex]) and with an application of de l'Hôpital's rule.

Choose [tex]$x,y\geq 0$[/tex].

If [tex]$p>0$[/tex], then [tex]$2\ (\min \{ x,y\})^p \leq x^p+y^p\leq 2\ (\max \{ x,y\})^p$[/tex], hence [tex]$\min \{ x,y\}\leq M_p(x,y)\leq \max \{ x,y\}$[/tex].
If [tex]$p=0$[/tex], then [tex]$(\min \{ x,y\})^2 \leq x\ y\leq (\max \{ x,y\})^2$[/tex], therefore [tex]$\min \{ x,y\} \leq G(x,y)=M_0(x,y)\leq \max \{ x,y\}$[/tex].
If [tex]$p<0$[/tex], then [tex]$2\ (\max \{ x,y\})^p\leq x^p+y^p\leq 2\ (\min \{ x,y\})^p$[/tex], so inequalities [tex]$\min \{ x,y\}\leq M_p(x,y)\leq \max \{ x,y\}$[/tex] follow.

Next, we show that relations a-c hold.
W.l.o.g. asume [tex]$x < y$[/tex], so that [tex]$x = \min \{ x,y\}$[/tex] and [tex]$y = \max \{ x,y\}$[/tex]; then:

[tex]$M_p(x,y)=y\ \left\{ \frac{1}{2}\ [1+(\tfrac{x}{y})^p]\right\}^\frac{1}{p} = x\ \left\{ \frac{1}{2}\ [1+(\tfrac{y}{x})^p]\right\}^\frac{1}{p}$[/tex]

and passing to the limit yields:

[tex]$\lim_{p\to +\infty} M_p(x,y) = y\ \lim_{p\to +\infty} \left\{ \frac{1}{2}\ [1+(\tfrac{x}{y})^p]\right\}^\frac{1}{p} =y=\max \{ x,y\}$[/tex]

[tex]$\lim_{p\to -\infty} M_p(x,y) = x\ \lim_{p\to -\infty} \left\{ \frac{1}{2}\ [1+(\tfrac{y}{x})^p]\right\}^\frac{1}{p} =x=\min \{ x,y\}$[/tex]

and a-b follow. To prove c we write:

[tex]$M_p(x,y) =\exp \left( \frac{1}{p}\ \ln \frac{x^p+y^p}{2}\right)$[/tex]

and evaluate:

[tex]$\lim_{p\to 0} \frac{1}{p}\ \ln \frac{x^p+y^p}{2} = \lim_{p\to 0} \frac{\ln \tfrac{1}{2}(x^p+y^p)}{\tfrac{1}{2}(x^p+y^p) -1}\ \frac{(x^p+y^p)-2}{2p}$[/tex]
[tex]$=\lim_{p\to 0} \frac{\ln \tfrac{1}{2}(x^p+y^p)}{\tfrac{1}{2}(x^p+y^p) -1}\ \frac{1}{2}\ \left[ \frac{x^p-1}{p} +\frac{y^p-1}{p}\right]$[/tex]
[tex]$=\lim_{p\to 0} \frac{\ln \tfrac{1}{2}(x^p+y^p)}{\tfrac{1}{2}(x^p+y^p) -1}\cdot \frac{1}{2}\ \left[ \lim_{p\to 0}\frac{x^p-1}{p} +\lim_{p\to 0}\frac{y^p-1}{p}\right]$[/tex]
[tex]$=1\cdot \frac{1}{2}\ (\ln x+\ln y)=\ln \sqrt{x\ y}$[/tex],

therefore:

[tex]$\lim_{p\to 0} M_p(x,y) =\exp \left( \lim_{p\to 0} \frac{1}{p}\ \ln \frac{x^p+y^p}{2} \right) =\sqrt{x\ y}=G(x,y)$[/tex]

as we claimed.

The last assertion is trivial.

[gugo82]

"gugo82":2. Let [tex]$x,y\geq 0$[/tex], choose [tex]$p\in \mathbb{R} \setminus \{ 0\}$[/tex] and set:

[tex]$M_p(x,y):=\left\{ \frac{1}{2}(x^p+y^p)\right\}^\frac{1}{p}$[/tex];

we say that [tex]$M_p(x,y)$[/tex] is the [tex]$p$[/tex]-mean of [tex]$x$[/tex] and [tex]$y$[/tex].

Using Lagrange multipliers method, show that for [tex]$p>q$[/tex] the following general inequality holds:

(pMqM) [tex]$M_p(x,y)\geq M_q(x,y)$[/tex]

and that equality case occurs in (pMqM) iff [tex]$x=y$[/tex] (in the case [tex]$p>0$[/tex]) or iff [tex]$x=y \text{ or } x=0 \text{ or } y=0$[/tex] (for [tex]$p<0$[/tex]).

a. Let [tex]$p>q>0$[/tex].

Since [tex]$M_q$[/tex] maps [tex]$[0,+\infty[^2$[/tex] onto [tex]$[0,+\infty[$[/tex], to prove (pMqM) it suffices to show that forall [tex]$a\geq 0$[/tex] the following inequality holds:

[tex]$\min_{\{ (x,y):\ M_q(x,y)=a\}} M_p(x,y) \geq a$[/tex].

Hence we're naturally led to solve the following constrained minimum problem:

(*) [tex]$\begin{cases} \text{minimize } M_p(x,y) \\ \text{u.c. } M_q(x,y)=a\end{cases}$[/tex],

which has actually a solution (for the objective function is continuous and the constraint is compact) and can be solved using Lagrange multipliers method; instead of solving (*) we prefer to solve the following equivalent problem:

(**) [tex]$\begin{cases} \text{minimize } 2M_p^p(x,y) \\ \text{u.c. } 2M_q^q(x,y)=2a^q\end{cases} \quad \Leftrightarrow \quad \begin{cases} \text{minimize } x^p+y^p \\ \text{u.c. } x^q+y^q=2a^q\end{cases}$[/tex]

in order to avoid dumb computations.

- Set [tex]$a>0$[/tex].
Lagrange equations associated with the problem (**) are:

[tex]$\begin{cases} px^{p-1}+\lambda \ qx^{q-1} =0\\ py^{p-1}+\lambda \ qy^{q-1} =0\\ x^q+y^q-2a^q=0\end{cases}$[/tex];

from the third equation we get [tex]$x,y\neq 0$[/tex] and first two we get:

[tex]$x=(-q\lambda)^\frac{q}{p-q}=y$[/tex]

which can be plugged into the last equation to find the right value of [tex]$\lambda$[/tex], i.e.:

[tex]$\lambda =-\frac{1}{q} \ a^{p-q}$[/tex].

Finally the unique stationary point of the objective function [tex]$2M_p^p(x,y)$[/tex] on the constraint [tex]$2M_q^q(x,y)=2a^q$[/tex] is [tex]$(x_a,y_a)=(a,a)$[/tex].

Next we show that the stationary point [tex]$(a,a)$[/tex] is actually a minimum using the Hessian test (as cited in my previous post).
The Hessian matrix [tex]$H(x,y)$[/tex] of the objective function in [tex]$(a,a)$[/tex] is:

[tex]$H(a,a):=\begin{pmatrix} p(p-1)\ x^{p-2} & 0\\ 0 & p(p-1)\ y^{p-2}\end{pmatrix}_{(x,y)=(a,a)} = \begin{pmatrix} p(p-1)\ a^{p-2} & 0\\ 0 & p(p-1)\ a^{p-2}\end{pmatrix} =p(p-1)a^{p-2} \ I$[/tex],

where [tex]$I$[/tex] is the [tex]$2\times 2$[/tex] identity matrix; the tangent vector space [tex]$\mathcal{T}_{(a,a)}$[/tex] to the constraint in [tex]$(a,a)$[/tex] is the set of vectors which are orthogonal to the gradient [tex]$D\left( 2M_q^q\right)(a,a)$[/tex]*, i.e.:

[tex]$\mathcal{T}_{(a,a)} := \left\{ u=(u_1,u_2) \in \mathbb{R}^2:\ u_1\ qa^{q-1}+u_2\ qa^{q-1} =0\right\} =\{ (u_1,-u_1)\}_{u_1\in \mathbb{R}}$[/tex];

the quadratic form associated to [tex]$H(a,a)$[/tex] in [tex]$\mathcal{T}_{(a,a)}$[/tex] is given by:

[tex]$\langle u, H(a,a)\ u\rangle =p(p-1) a^{p-2}\ |u|^2$[/tex]

and it's obviously positive definite.
Therefore [tex]$(a,a)$[/tex] is a local minimum of [tex]$2M_p^p(x,x)$[/tex] on the constraint [tex]$2M_q^q(x,y)=2a^q$[/tex], and we have [tex]$2M_p^p(a,a)=2a^p$[/tex].
We claim that [tex]$(a,a)$[/tex] is a global minimum. Since the objective function and the constraint are smooth, there are no other points on the constraint which can give the minimum to [tex]$2M_p^p(x,y)$[/tex] except the stationary point [tex]$(a,a)$[/tex] and the extremal points of the constraint itself, i.e. the points [tex]$(2^{\tfrac{1}{q}}\ a,0)$[/tex] and [tex]$(0,2^{\tfrac{1}{q}}\ a)$[/tex]; hence it suffices to evaluate [tex]$2M_p^p (2^{\tfrac{1}{q}}\ a,0),\ 2M_p^p (0,2^{\tfrac{1}{q}}\ a)$[/tex] to prove our claim.
But [tex]$2M_p^p(2^{\tfrac{1}{q}}\ a,0)=2M_p^p(0,2^{\tfrac{1}{q}}\ a)=2^{\tfrac{p}{q}}\ a^p \geq 2a^p=2M_p^p(a,a)$[/tex] and [tex]$(a,a)$[/tex] is a global minimum.

- On the other hand, if we set [tex]$a=0$[/tex], then the constraint reduces to [tex]$\{ (0,0)\}$[/tex] and trivially [tex]$\min_{\{ (x,y):\ 2M_q^q(x,y)=0\}} 2M_p^p(x,y)=0$[/tex].

Finally, the original problem (*) has the solution [tex]$(a,a)$[/tex] and [tex]$\min_{\{ (x,y):\ M_q(x,y)=a\}} M_p(x,y)=a$[/tex] as we wanted.


________
* We have [tex]$D(2M_q^q)(x,y)=(qx^{q-1},qy^{q-1})$[/tex].

b. Let [tex]$p>0>q$[/tex].

In this case also we're led to the constrained minimum problem (**), but the constraint is no longer compact; therefore we have to check more carefully the existence of a solution to (**).
The objective function [tex]$2M_p^p(x,y)$[/tex] is continuous and goes to [tex]$+\infty$[/tex] as [tex]$|(x,y)| \to +\infty$[/tex], hence [tex]$2M_p^p$[/tex] possesses a global minimum on the constraint.

- Set [tex]$a>0$[/tex].
We can repeat the whole set of computations previously made to obtain that [tex]$(a,a)$[/tex] is a stationary point and a local minimum of [tex]$2M_p^p$[/tex] on the constraint; moreover [tex]$2M_p^p(a,a)=2a^p$[/tex].
In order to show that [tex]$(a,a)$[/tex] is actually a global minimum for the objective function, we reason as follows.

Since [tex]$2M_p^p(x,y)$[/tex] diverges when [tex]$|(x,y)|$[/tex] increases, there exists [tex]$R>0$[/tex] so large that [tex]$2M_p^p(x,y) \geq 3a^p$[/tex] for all [tex]$(x,y)\in [0,+\infty[^2\setminus \mathcal{B} (\mathfrak{o} ,R)$[/tex]; in particular we have [tex]$2M_p^p(x,y) \geq 3a^p$[/tex] for all [tex]$(x,y)\in \{ 2M_q^q(x,y)=2a^q\} \setminus \mathcal{B} (\mathfrak{o} ,R)$[/tex].
Now consider the set [tex]$K_R:=\{ 2M_q^q(x,y)=2a^q\} \cap \overline{\mathcal{B}} (\mathfrak{o} ,R)$[/tex]: this set is compact, hence the objective function attains a minimum in [tex]$K_R$[/tex] (by Weierstrass Theorem). Since [tex]$(a,a)\in K_R$[/tex], we have [tex]$0\leq \min_{K_R} 2M_p^p \leq 2a^p$[/tex]; furthermore the minimum cannot be attained in other points except [tex]$(a,a)$[/tex] or the two extremal points of [tex]$K_R$[/tex], say [tex]$(x_1,y_1),\ (x_2,y_2)$[/tex].
Now we can estimate [tex]$2M_p^p(x_i,y_i)$[/tex] using continuity: in fact if we choose a sequence [tex]$(x_i^n,y_i^n) \in \{ 2M_q^q(x,y)=2a^q\} \setminus \mathcal{B} (\mathfrak{o} ,R)$[/tex] s.t. [tex]$(x_i^n,y_i^n)\to (x_i,y_i)$[/tex] we get:

[tex]$2M_p^p(x_i,y_i)=\lim_n 2M_p^p(x_i^n,y_i^n) \geq 3a^p$[/tex]

hence [tex]$2M_p^p$[/tex] attains its minimum on [tex]$K_R$[/tex] in [tex]$(a,a)$[/tex].

Finally, [tex]$(a,a)$[/tex] is the global minimum of [tex]$2M_p^p$[/tex] on the constraint [tex]$2M_q^q(x,y)=2a^q$[/tex]: in fact suppose it were not the case so that there exists at least a point [tex]$(\xi ,\eta) \in \{ 2M_q^q(x,y)=2a^q\}$[/tex] s.t. [tex]$2M_p^p(\xi ,\eta)<2a^p=2M_p^p(a,a)$[/tex]; but [tex]$(\xi ,\eta)$[/tex] cannot belong to [tex]$\{ 2M_q^q(x,y)=2a^q\} \setminus \mathcal{B} (\mathfrak{o} ,R)$[/tex] (because of the very definition of [tex]$R$[/tex]), hence [tex]$(\xi ,\eta )\in K_R$[/tex] and therefore [tex]$2M_p^p(\xi ,\eta)\geq 2a^p$[/tex]. Contraddiction!

- On the other hand, if we set [tex]$a=0$[/tex], the constraint reduces to the union of the nonnegative [tex]$x$[/tex] and [tex]$y$[/tex] semiaxes, i.e. to the set [tex]$\{ (x,0)\}_{x\geq 0} \cup \{ (0,y)\}_{y\geq 0}$[/tex].
In this case it's easily seen that [tex]$\min_{\{ 2M_q^q (x,y)=0\}} 2M_p^p(x,y) =2M_p^p(0,0)=0$[/tex].

It then follows that [tex]$(a,a)$[/tex] is the unique solution of (*) and that [tex]$\min_{\{ M_q (x,y)=a\}} M_p(x,y) =a$[/tex].

c. The case [tex]$0>p>q$[/tex] can be handled in a similar way.

Therefore (pMqM) holds. The characterization of the equality cases is a staightforward consequence of what we've already proved.

[dissonance]

This one is very interesting. Using a parametrization for [tex]Z[/tex] is really a cheap trick, since it voids the necessity to use any advanced technique such as Lagrange multipliers. That's why I didn't like my solution. I'll memorize this method, thank you.

Ok, now for pMqM. I'll skip directly to the "elementary calculus" proof for a change.
Let [tex]p > q[/tex]. As you pointed out in your "Hints" box, the inequality

[tex]M_p(x, y) \ge M_q(x, y),\ \mathrm{all}\ x, y>0[/tex];

is the same as

[tex]M_p(t, 1) \ge M_q (t, 1),\ \mathrm{all}\ t>0[/tex];

so we're going to prove this last one, which is best written

[tex](t^q +1) ^ {1 \over q} ( t^p +1 )^ {-{1 \over p}} \le 2 ^{{1\over q} - {1 \over p}},\ \mathrm{all}\ t>0[/tex].

Let [tex]\varphi(t)=(t^q +1) ^ {1 \over q} ( t^p +1 )^ {-{1 \over p}},\ t>0[/tex]. This function tends to [tex]1[/tex] as [tex]t \to 0, t \to +\infty[/tex] and attains the value [tex]2 ^{{1\over q} - {1 \over p}}[/tex] for [tex]t=1[/tex]; we claim that this is its only extreme point and hence its global maximum ([tex]1 < 2 ^{{1\over q} - {1 \over p}}[/tex] because of the conditions on [tex]p[/tex] and [tex]q[/tex]) .

This can be proved via the following brute-force computations (which almost drove me to the asylum ):

[tex]\varphi'(t)=(t^q+1)^{-{q-1 \over q }} t^{q-1} (t^p +1 ) ^{-{1 \over p}} - (t^q +1)^{1 \over p} (t^p +1 )^{-{1+p \over p}}t^{p-1}[/tex];

having assumed that [tex]t>0[/tex] we can multiply l.h.s and r.h.s. of the equation [tex]\varphi'(t)=0[/tex] by [tex](t^q +1)^{-1 \over p} (t^p +1 )^{{1+p \over p}}t^{-p+1}[/tex] and get the equivalent

[tex]\displaystyle {t^{-p+q} -1 \over t^q +1}=0[/tex]

which clearly has the only positive zero [tex]t=1[/tex]. This proves also the equality case (except for negative [tex]p[/tex] - that's trivial enough to leave unproved).

It's a curious thing, that at the beginning had confused me, that for [tex]p > q \ge 1[/tex] the inequalities between the [tex]p[/tex]-norms (which differ from [tex]p[/tex]-means by a constant) go exactly in the opposite direction.

[gugo82]

Ok!

You used a parametrization to check that $(1,1)$ is indeed a minimum... But there are other (not so easy) ways!

For example, one could also check the sufficient condition with the Hessian matrix (which is just a suitable variant of the classical unconstrained extremum sufficient condition) in order to prove that a stationary point on the constraint is a local minimum.
In fact the following holds:

In order to find a strict local minimum [maximum] for a [tex]$C^2$[/tex] function [tex]$f(x,y)$[/tex] in a stationary point [tex]$(x_0,y_0)$[/tex] on the constraint [tex]$Z:g(x,y)=0$[/tex], it is sufficient the (quadratic form associated to the) Hessian matrix [tex]$H_f(x_0,y_0)$[/tex] to be positive [negative] definite in the tangent space [tex]$\mathcal{T}_Z(x_0,y_0)$[/tex] to [tex]$Z$[/tex] in [tex]$(x_0,y_0)$[/tex].

In our context this condition fails, for the Hessian matrix of [tex]$f(x,y)=A(x,y)$[/tex] is null.
Neverthless one can take [tex]$f(x,y)=G^2(x,y)$[/tex], [tex]$g(x,y)=2A(x,y)-2$[/tex] (the square in [tex]$f$[/tex] and the scaling factor in [tex]$g$[/tex] are chosen to avoid dumb computations) in order to transform the minimum problem into a maximum problem: the sufficient condition works in this case, because [tex]$H_f(1,1)$[/tex] is non null.

[dissonance]

Here's my try at 1).
Let's start with AMGM. First of all we point out that the case [tex]x=0[/tex] or [tex]y=0[/tex] is trivial; in what follows we assume [tex]x, y>0[/tex].
Both [tex]A[/tex] and [tex]G[/tex] are positively homogeneous of the same degree. So to prove AMGM it will suffice to prove that [tex]G(x, y)=1 \Rightarrow A(x, y) \ge 1[/tex]; from this we will get

[tex]\displaymath A(\frac{x}{G(x, y)}, \frac{y}{G(x, y)}) \ge G(\frac{x}{G(x, y)}, \frac{y}{G(x, y)})[/tex]

and the claim follows by homogeneity.

Since [tex]\frac{\partial G}{\partial x}=\frac{y}{2\sqrt{xy}}[/tex] and [tex]\frac{\partial G}{\partial y}=\frac{x}{2\sqrt{xy}}[/tex], every value of [tex]G[/tex] other than [tex]0[/tex] is regular; hence we can find every extreme point of [tex]A[/tex] on [tex]Z=\{(x, y) \mid G(x, y)=1\}[/tex] by means of Lagrange multipliers. Turns out that there is only one such point: [tex](x, y)=(1, 1)[/tex]; that's because

[tex]\displaymath \begin{cases}
\frac{\partial {A}}{\partial{x}}-\lambda \frac{\partial{G}}{\partial{x}} &=0 \\
\frac{\partial {A}}{\partial{y}}-\lambda \frac{\partial{G}}{\partial{y}} &=0 \\
G(x, y) &=1 \\ \end{cases}[/tex]

has the only solution [tex]x=y=\lambda=1[/tex], as it is readily seen.

Finally, we show that [tex]A[/tex] has got a global minimum on [tex]Z[/tex].
Here I couldn't think of anything better than use this parametrization of [tex]Z[/tex]:
[tex]Z=\{ (x, y) \mid y=1/x \}[/tex]
and the fact that [tex]\varphi(x)=A(x, 1/x)=1/2(x+1/x)[/tex] is positive and infinite for [tex]x \to 0,\ x \to +\infty[/tex]. From this it follows that [tex]A[/tex] must have a global minimum on [tex]Z[/tex]; and since it has only one extreme point, that one must be it.
This proves AMGM.

Now for the other two inequalities. Since the technique is essentially the same I will cover them quickly. For AMHM we prove that [tex]A(x, y)=1 \Rightarrow 1 \ge H(x, y)[/tex] and use homogenity as before; again [tex]1[/tex] is a regular value for [tex]A[/tex] and so every extreme point of [tex]H[/tex] over [tex]Z=\{ (x, y)\mid A(x, y)=1 \}[/tex] will be detected via Lagrange multipliers, and again the only point as such is [tex](1, 1)[/tex].

The only difference from before is that we now need to prove this point is a global maximum for [tex]H[/tex] on [tex]Z[/tex]; to do so we write [tex]Z=\{ (x, y)\mid 0 < x < 2\ y=2-x \}[/tex] and evauate [tex]\varphi(x)=H(x, 2-x)=2x-x^2[/tex]; clearly [tex]x=1[/tex] is the global maximum of [tex]\varphi[/tex].
This proves AMHM.

GMHM goes the same way as AMHM: we prove that [tex]G(x, y)=1 \Rightarrow 1 \ge H(x, y)[/tex] by posing [tex]Z=\{(x, y) \mid G(x, y)=1 \}[/tex] and applying Lagrange method. Computations get fierce here, though.