If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.
 A: A geometric proof:
The equation $x+y+z=6$ describes a plane with intercepts 6 on all three axes.
The equation $x^2+y^2+z^2=12$ describes a sphere of radius $\sqrt{12}$ centered at the origin $x=y=z=0$. The inequality $x^2+y^2+z^2\ge12$ describes points not in the interior of this sphere.
It is easy to check that no points of the plane $x+y+z=6$ are in the interior of this sphere. (Use distance from point to plane. The plane's closest point to the origin is $x=y=z=2$, this point belongs to the sphere but is not inside the sphere. All other points of the plane are farther away from the origin.)
But that's precisely what we are asked to prove - if a point is in the plane, it's not inside the given sphere.
A: We need to prove that
$$a^2+b^2+c^2\geq12\left(\frac{a+b+c}{6}\right)^2$$ or
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
A: Proof:$$a^2+b^2\geq 2ab$$so:$$3(a^2+b^2+c^2)=a^2+b^2+c^2+2(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2=36$$Done!
A: Use Cauchy–Schwartz:
$$a+b+c\ \leqslant\ \sqrt{1^2+1^2+1^2}\sqrt{a^2+b^2+c^2}$$
A: Suppose $a+b$ is fixed then $$2(a^2+b^2)=(a+b)^2+(a-b)^2\ge (a+b)^2$$ with equality iff $a=b$
If any pair of $a,b,c$ are distinct we can reduce the sum of squares by replacing them by their average. The minimum value is therefore achieved when all three are equal.
Note that the condition that $a,b,c$ are non-negative is redundant.
The more general methods are more elegant, but such ascent/descent methods can be useful to see what is going on, and it is sometimes useful to reduce the number of variables you are dealing with.
