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thanks for your time. i am interested in various ways/techniques/tricks/methods (induction, convexity, concavity, maximum, minimum, geometry, trigonometry, ...) for proving the inequalities and their generalizations

(1) $\sqrt{1-x_1^2 -y_1^2} + \sqrt{1-x_2^2 -y_2^2} + \sqrt{1-x_3^2 -y_3^2} \le 3\,\sqrt{1-\left(\frac{x_1+x_2+x_3}{3}\right)^2 - \left(\frac{y_1+y_2+y_3}{3}\right)^2} $

i think for $0 \le x_1,y_1,z_1 \le 1$

(2) $\sqrt{x_1^2 + y_1^2} + \sqrt{x_2^2 + y_2^2} + \sqrt{x_3^2 + y_3^2} \ge 3 \sqrt{\left(\frac{x_1+x_2+x_3}{3}\right)^2 + \left(\frac{y_1+y_2+y_3}{3}\right)^2}$

for all $x_1,y_1,z_1$

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There's no $z$ in you formulas. Although I guess in each of them the particular combinations of $x$ and $y$ can be interpreted as a z-coordinate. What 1) is then saying is that if you take three points in a disk, the heights of the perpendiculars cut off by the sphere with equatorial plane that disk all sum to a number smaller than 3 times the height for the center of gravity of the three points. A similar in interpretation is possible for 2) except you won't have a sphere but a cone. – Raskolnikov Mar 26 '11 at 10:32
@Raskolnikov. Thanks. Thsese are definitely the geometrical interpretations. How about methods of proving them? – Mia Mar 26 '11 at 10:34
I suggest using concavity/convexity of the sphere/cone. Then it's just one line to prove them. (Unless you also need to prove that the sphere and cone have these properties to begin with.) – Raskolnikov Mar 26 '11 at 10:47
up vote 3 down vote accepted

For (1) I think you need to use the concavity of the function $f(t)=\sqrt{1-t^2}$, i.e. the inequality:

$$ f(\lambda_1 t_1+\lambda_2t_2+\lambda_3t_3) \geq \lambda_1 f(t_1) + \lambda_2 f(t_2)+ \lambda_3 f(t_3) $$

for $t_1,t_2,t_3\in [-1,1]$ and $\lambda_1,\lambda_2,\lambda_3\in [0,1]$ s.t. $\sum_{k=1}^3 \lambda_k=1$; and also the monotonicity of $f(t)$ in $[0,1]$.

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I do not understand what $t_i$ the function $f$ in your post should be applied to. Nor why the monotonicity of $f$ is needed. – Did Mar 31 '11 at 16:16
@Didier: Sorry, but I hadn't seen your comment until now. To answer your question: if I didn't miscalculate, $t_i=\sqrt{x_i^2+y_i^2}$ and $\lambda_i=3^{-1}$; the monotonicity applies when one tries to prove that $3f(\sum_i \lambda_i t_i)$ is actually less than or equal to the RHside of (1). – Pacciu Apr 2 '11 at 15:52
No problem, and thanks for the answer. The funny thing is that there exists a proof (of a more general statement) which uses no monotonicity at all, but only convexity. – Did Apr 2 '11 at 16:47

As mentioned by others, (1) and (2) are convexity/concavity inequalities, see this for example. A generalization of both inequalities which can help to see what is going on is as follows.

For every $z$ in $\mathbb{R}^n$, let $\|z\|$ denote its Euclidean norm. Consider $x_1$, $x_2$, ..., $x_K$ in $\mathbb{R}^n$, and introduce their barycenter $$ x=\frac1K\sum_{k=1}^Kx_k. $$ One sees that (1) is a special case of the fact that, if $\|x_k\|\le1$ for every $k$, then $$ \frac1K\sum_{k=1}^K\sqrt{1-\|x_k\|^2}\le \sqrt{1-\|x\|^2}. $$ Likewise, (2) is a special case of the fact that $$ \frac1K\sum_{k=1}^K\|x_k\|\ge \|x\|. $$ The OP asked about the case $n=2$ and $K=3$.

This general formulation indicates that these inequalities are true if (and only if) some specific functions $\varphi$ and $\psi$ defined on $\mathbb{R}^n$ are concave, respectively convex. Here, $$ \varphi(z)=\sqrt{1-\|z\|^2}, \quad \psi(z)=\|z\|. $$ A simple way to prove that a function is concave/convex is to write it as an infimum/supremum of affine functions. Recall that an affine function on $\mathbb{R}^n$ is defined by $z\mapsto\langle a,z\rangle+b$ for a given $a$ in $\mathbb{R}^n$ and a given $b$ in $\mathbb{R}$. Now, $$ \varphi(z)=\inf_{u\in\mathbb{R}^n}\langle u,z\rangle+\sqrt{1+\|u\|^2},\quad \psi(z)=\sup_{\|u\|=1}\langle u,z\rangle, $$ hence the generalizations of (1) and (2) hold.

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$\varphi (z)$ is the Legendre transform of $-\sqrt{1+\| u\|^2}$, right? – Pacciu Apr 2 '11 at 17:18
@Pacciu: Nope, $-\varphi$ is the Legendre transform of $u\mapsto\sqrt{1+\|u\|^2}$. – Did Apr 2 '11 at 18:30

The proof of the second inequality: Let $r_1=[x_1,y_1]^T, r_2=[x_2,y_2]^T, r_3=[x_3,y_3]^T$ be three vectors in $\Re ^2$. Then the second inequality is equivalent to the following one. $$\|r_1\|_2+\|r_2\|_2+\|r_3\|_2 \ge \|r_1+r_2+r_3\|_2$$ This inequality is obvious. It is known as the triangle inequality or Cauchy-Schwarz inequality.

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Cauchy-Schwarz inequality is NOT triangle inequality. – Did Sep 16 '11 at 5:35

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