# Solve a^2+b^2+c^2>= a+b+c ?

Feb 1, 2017

See below.

#### Explanation:

${a}^{2} + {b}^{2} + {c}^{2} \ge a + b + c \to {a}^{2} + {b}^{2} + {c}^{2} - a - b - c \ge 0$

So we can arrange this as an optimization problem

Find

${\left(a , b , c\right)}_{\circ} = \text{argmin} \left({a}^{2} + {b}^{2} + {c}^{2} - a - b - c\right)$

subjected to $a b c = 1$

If this solution gives

${a}_{\circ}^{2} + {b}_{\circ}^{2} + {c}_{\circ}^{2} - {a}_{\circ} - {b}_{\circ} - {c}_{\circ} \ge 0$ then the question is proved.

To prove this we will use the Lagrange Multipliers technique. The lagrangian is

$L \left(a , b , c , \lambda\right) = {a}^{2} + {b}^{2} + {c}^{2} - a - b - c + \lambda \left(a b c - 1\right)$

The stationary point conditions are

{(2 a-1 + b c lambda=0),( 2 y-1 + a c lambda=0), (2 c-1 + a b lambda=0),( a b c-1=0):}

or

{(2 a-1 + lambda/a=0),( 2 b-1 + lambda/b=0), (2 c-1 + lambda/c=0),( a b c-1=0):}

which gives

$a = b = c = \frac{1}{4} \left(1 \pm \sqrt{1 - 8 \lambda}\right)$

or

${\left(\frac{1}{4} \left(1 \pm \sqrt{1 - 8 \lambda}\right)\right)}^{3} = 1 \to \frac{1}{4} \left(1 \pm \sqrt{1 - 8 \lambda}\right) = 1$

giving $\lambda = - 1$

The solution is ${a}_{\circ} = {b}_{\circ} = {c}_{\circ} = 1$

This is a minimum point because the hessian of ${a}^{2} + {b}^{2} + {c}^{2} - a - b - c$ is $\left(\begin{matrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{matrix}\right)$ which is definite positive.

The minimum value is $0$. So it is proved