Inequality: Prove $a^4+b^4+c^4 \ge abc(a+b+c)$

[kaliprasad]

for positive a , b, c prove that $a^4+b^4+c^4 \ge abc(a+b+c)$

 

[Albert1]

for positive a , b, c prove that $a^4+b^4+c^4 \ge abc(a+b+c)$

Spoiler

Using :$AM\ge GM$
$a^4+b^4+c^4 \ge a^2b^2+b^2c^2+c^2a^2\ge a^2bc+b^2ca+c^2ab=abc(a+b+c)$
equality holds when :$a=b=c$

 

[kaliprasad]

above solution is right

my full solution ( same as above)

Spoiler

this can be done in 2 steps

we know by AM GM inequality

$a^4+b^4 \ge 2 a^2b^2$
$b^4+c^4 \ge 2 b^2 c^2$
$c^4 + a^4 \ge 2 a^2b^2$

adding the 3 above and dividing by 2 we get

$a^4+b^4+c ^4 >= a^2b^2+b^2c^2 + c^2 a^2 \cdots 1$

now we repeat the process with $a^2b^2$ , $b^2 c^2$ and $c^2 a^2$ to get as below

$a^2 b^2 + b^2 c^2 > = 2 b^2ac$
$b^2c^2 + c^2 a^2 >= 2 c^2ab$
$c^2a^2 + a^2 b^2 >= 2 a^2bc$

adding the above and dividing by 2 we get

$a^2b^2 + b^2 c^2 + c^2 a^2 \ge (b^2ac+c^2ab+a^2bc)$ or $abc(b+c+a)\cdots 2$

from (1) and (2) it follows

$a^4 + b^4 + c^4 >= abc(a+b+c)$