Given that $5a+2b+3c=10$, What is the minimum value of $a^2+b^2+c^2$? The question is , 
Given that $$5a+2b+3c=10$$
What is the minimum value of $$a^2+b^2+c^2$$?
I know that I have to use AM-GM inequality somehow but I have no idea how to use it for this problem. Help would be much appreciated.
 A: This problem can be solved in many ways.
I'm gonna solve it using vectors.
Writing  $5a+2b+3c$ as the dot product of 2 vectors,
$$(5\hat{i} +2\hat {j}+3\hat k)\cdot(a\hat i+b\hat j+c\hat k)=10$$ 
Recall that, by definition , $\vec A.\vec B=|\vec A|\ |\vec B|\cos\theta$ where $\theta$ is the angle between the vectors.
So, $$10=|5\hat i +2\hat{j}+3\hat k|\ |a\hat i+b\hat{j}+c\hat k|\cos\theta$$
$$10=\sqrt{5^2+2^2+3^2}\cdot\sqrt{a^2+b^2+c^2}\cdot\cos\theta$$
Squaring,$$100=38\cdot\big(a^2+b^2+c^2\big)\cdot \cos^2\theta $$
$$\dfrac{100}{38}\cdot \sec^2\theta=\big(a^2+b^2+c^2\big)$$
but , the minimum value of $\sec^2\theta=1$, so the minimum value of $\big(a^2+b^2+c^2\big)$ is $\dfrac{100}{38}$. 
You're done!!
A: I'm going to use $x,y,z$ instead of $a,b,c$. Obviously this doesn't change the problem.
Let $\mathbf n = (5,2,3)/\sqrt{38}$. The condition $5x + 2y + 3z = 10$ can then be written as $\mathbf n \cdot (x,y,z) = 10/\sqrt{38}$. This is the equation of a plane with unit normal vector $\mathbf n$ at a distance $10/\sqrt{38}$ from the origin. Finding the minimum value of $x^2 + y^2 + z^2$ on this plane means we have to minimize the squared distance from the plane to the origin. 
The distance from the plane to the origin is $10/\sqrt{38}$, so the squared distance is $100/38$.
