Find the minimum a^4+b^4+c^4−3abc

[lfdahl]

Find the minimum of the expression:

$$a^4+b^4+c^4-3abc$$

- if $a,b$ and $c$ are real numbers satisfying the conditions: $a \ge 1$ and $a+b+c = 0$.

Hint:

Spoiler

First prove the two auxiliary inequalities:

1. $bc \le \frac{a^2}{4}.$

2. $b^4+c^4 \ge \frac{a^4}{8}.$

 

[Albert1]

Find the minimum of the expression:

$$a^4+b^4+c^4-3abc=A$$

- if $a,b$ and $c$ are real numbers satisfying the conditions: $a \ge 1$ and $a+b+c = 0$.

Hint:

Spoiler

First prove the two auxiliary inequalities:

1. $bc \le \frac{a^2}{4}---(1)$

2. $b^4+c^4 \ge \frac{a^4}{8}---(2)$

my solution(using hint):

Spoiler

proof of (1):
$2bc=(b+c)^2-(b^2+c^2)\leq a^2-2bc$
so $4bc\leq a^2$
proof of (2):
$b^4+c^4=(b^2+c^2)^2-2b^2c^2=((b+c)^2-2bc)^2-2b^2c^2$
$=(a^2-2bc)^2-2b^2c^2\geq\dfrac {2a^4-a^4}{8}=\dfrac{a^4}{8}$
$A\geq a^4+\dfrac {a^4}{8}-3abc\geq \dfrac {9a^4}{8}-\dfrac {3a^3}{4}$
$\geq \dfrac{3a^4}{8}\geq\dfrac{3}{8}$
with ($a=1,b=c=\dfrac{-1}{2}$)

 

[lfdahl]

my solution(using hint):

Spoiler

proof of (1):
$2bc=(b+c)^2-(b^2+c^2)\leq a^2-2bc$
so $4bc\leq a^2$
proof of (2):
$b^4+c^4=(b^2+c^2)^2-2b^2c^2=((b+c)^2-2bc)^2-2b^2c^2$
$=(a^2-2bc)^2-2b^2c^2\geq\dfrac {2a^4-a^4}{8}=\dfrac{a^4}{8}$
$A\geq a^4+\dfrac {a^4}{8}-3abc\geq \dfrac {9a^4}{8}-\dfrac {3a^3}{4}$
$\geq \dfrac{3a^4}{8}\geq\dfrac{3}{8}$
with ($a=1,b=c=\dfrac{-1}{2}$)

Thankyou, Albert! - for your nice solution. Well done!