# Thread: (a^2+2)(b^2+2)(c^2+2)>= 3(a+b+c)^2

1. ## (a^2+2)(b^2+2)(c^2+2)>= 3(a+b+c)^2

Prove that $\displaystyle \forall a, b, c\in\mathbb{R}$: $\displaystyle (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$.

2. Originally Posted by james_bond
Prove that $\displaystyle \forall a, b, c\in\mathbb{R}$: $\displaystyle (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$.
Let me take a stab at this. Let's define the function F as

$\displaystyle F(a,b,c) = (a^2+2)(b^2+2)(c^2+2) - 3(a+b+c)^2$

We will show that the minumum of this is zero. Taking partial derivatives and setting to zero gives

$\displaystyle F_a = 2a(b^2+2)(c^2+2) - 6(a+b+c) = 0$
$\displaystyle F_b = 2b(a^2+2)(c^2+2) - 6(a+b+c) = 0$
$\displaystyle F_c = 2c(a^2+2)(b^2+2) - 6(a+b+c) = 0,$

By substracting in pairs we obtain

$\displaystyle (a-b)(ab-2) = 0,\;\;\;(a-c)(ac-2) = 0,\;\;\;(b-c)(bc-2) = 0$

from which we deduce that $\displaystyle a = b = c.$

Substituting into any of the partial derivatives gives

$\displaystyle 2a(a-1)(a+1)(a^2+5) = 0$

giving rise to three cases: $\displaystyle a = 0, \;\;a = -1, \;\;a=1.$

The first gives $\displaystyle F(0,0,0) = 8$ whereas the remaining two gives $\displaystyle F(- 1,- 1,- 1) = 0,\;\;\;F(1, 1, 1) = 0$. Thus, the minimum of F is 0 giving

$\displaystyle F(a,b,c) \ge 0$

for which the inequality follows.

3. Originally Posted by james_bond

Prove that $\displaystyle \forall a, b, c\in\mathbb{R}$: $\displaystyle (a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$.
changing $\displaystyle a,b,c$ to $\displaystyle \sqrt{2}a, \sqrt{2}b, \sqrt{2}c$ respectively, we'll get the following equivalent inequality: $\displaystyle 4(a^2 + 1)(b^2 + 1)(c^2 + 1) \geq 3(a+b+c)^2. \ \ \ \ \ (*)$

Lemma: $\displaystyle \forall a,b,c \in \mathbb{R}: \ \ a^2 + b^2 + c^2 + 1 + 4a^2b^2c^2 - 2(ab+bc+ac) \geq 0.$

Proof: rewrite the inequality as: $\displaystyle (1+4b^2c^2)a^2 - 2(b+c)a + (b-c)^2 + 1 \geq 0.$ now look at the LHS of the inequality as a quadratic function in $\displaystyle a.$

the discriminant of this quadratic function is: $\displaystyle -4[(2bc - 1)^2 + 4b^2c^2(b-c)^2] \leq 0. \ \ \Box$

Proof of $\displaystyle (*)$: $\displaystyle 4(a^2 + 1)(b^2+1)(c^2 + 1) - 3(a+b+c)^2 =$

$\displaystyle (2ab \ - \ 1)^2 + (2bc \ - \ 1)^2 + (2ac \ - \ 1)^2 +a^2 + b^2 + c^2 + 1 + 4a^2b^2c^2 - 2(ab \ + \ bc \ + \ ac) \geq 0,$ by the Lemma. $\displaystyle \Box$

4. Originally Posted by danny arrigo
$\displaystyle (a-b)(ab-2) = 0,\;\;\;(a-c)(ac-2) = 0,\;\;\;(b-c)(bc-2) = 0$

from which we deduce that $\displaystyle a = b = c.$
What if $\displaystyle a=\frac 2b=c$?

5. Originally Posted by james_bond
What if $\displaystyle a=\frac 2b=c$?
Yes, you are correct, they are solutions of my intermediate set of equations but will not satisfy the original set. Notice that

$\displaystyle a = 0,\;\;b = 0,\;\;c = 0$

is a solution of the original set but cannot be obtained from your result.

2>, 2b2, 2c2