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

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The discussion revolves around proving the inequality involving distances in three-dimensional space. Participants are asked to provide a solution without using advanced concepts like norms, aiming for a high school level approach. A reminder is given about the forum's expectation that members should present their own solutions when posting problems. The conversation emphasizes clarity in mathematical reasoning and encourages sharing simpler methods for proving inequalities. The focus remains on finding accessible solutions to the stated inequality.
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Prove:

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

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

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
 
My solution:

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}
 
I like Serena said:
My solution:

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
 
solakis said:
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?
 
A high school solution is the following:
$$\sqrt{a^2+b^2+c^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}+\sqrt{x^2+y^2+z^2}$$ $$\Longleftrightarrow\sqrt{a^2+b^2+c^2}-\sqrt{x^2+y^2+z^2}\leq\sqrt{(x-a)^2+(y-b)^2+(z-c)^2} $$

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
 
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