Inequality c≤√[(x−a^2+(y−b)^2+(z−c)^2]+√(x2+y2+z2)

[solakis1]

Prove:

$$\sqrt{a^2+b^2+c^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}+\sqrt{x^2+y^2+z^2}$$

 

[kaliprasad]

Prove:

$$\sqrt{a^2+b^2+c^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}+\sqrt{x^2+y^2+z^2}$$

Spoiler

This comes from law of triangle sides in 3 dimensions

distance from origin to (a,b,c) < distance from (x,y,z) to (a,b,c) + distance from origin to (x,y,z) and equal if (x,y,z) is between (0,0) and (a,b,c ) and in the line between the 2

 

[I like Serena]

My solution:

Spoiler

Avoiding geometry notions, and instead using only the axioms of norm from linear algebra, and the fact that the Euclidean norm is actually a norm.

Let $\mathbf a = (a,b,c)$ and $\mathbf x = (x,y,z)$.
Then:
\begin{aligned}\sqrt{a^2+b^2+c^2} = \|\mathbf a\| = \|(\mathbf a - \mathbf x) + \mathbf x\|
\le \|\mathbf a - \mathbf x\| + \|\mathbf x\|
&= \|\mathbf x - \mathbf a\| + \|\mathbf x\| \\
&= \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2} + \sqrt{x^2+y^2+z^2}
\end{aligned}

 

[solakis1]

My solution:

Spoiler

Avoiding geometry notions, and instead using only the axioms of norm from linear algebra, and the fact that the Euclidean norm is actually a norm.

Let $\mathbf a = (a,b,c)$ and $\mathbf x = (x,y,z)$.
Then:
\begin{aligned}\sqrt{a^2+b^2+c^2} = \|\mathbf a\| = \|(\mathbf a - \mathbf x) + \mathbf x\|
\le \|\mathbf a - \mathbf x\| + \|\mathbf x\|
&= \|\mathbf x - \mathbf a\| + \|\mathbf x\| \\
&= \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2} + \sqrt{x^2+y^2+z^2}
\end{aligned}

Without using the norm what solution would you suggest, i mean a high school solution

 

[MarkFL]

Without using the norm what solution would you suggest, i mean a high school solution

When you post a problem here in our "Challenge Questions and Puzzles" forum, it is expected that you have a solution to the problem already (read http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html)...what solution do you have?

 

[solakis1]

A high school solution is the following:

Spoiler

$$\sqrt{a^2+b^2+c^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}+\sqrt{x^2+y^2+z^2}$$ $$\Longleftrightarrow\sqrt{a^2+b^2+c^2}-\sqrt{x^2+y^2+z^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2} $$

And by squaring and cancelling equal terms on both sides repeatedly we end up with the equivalent formula:

$$(a^2y^2+b^2x^2-2axby)+(a^2z^2+c^2x^2-2axzc)+(b^2z^2+c^2y^2-2byzc)\geq 0$$, which is true hence the initial formula is true