MHB How can positive numbers be used to prove an inequality challenge?

[anemone]

If $a,\,b,\,c$ are positive numbers, show that $8(a^3+b^3+c^3)\ge (a+b)^3+(a+c)^3+(b+c)^3$.

 

[Albert1]

If $a,\,b,\,c$ are positive numbers, show that $8(a^3+b^3+c^3)\ge (a+b)^3+(a+c)^3+(b+c)^3----(1)$.

Spoiler

if (1) is true then expanding and simplifying :
$8(a^3+b^3+c^3)\ge 2a^3+2b^3+2c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2$
$2(a^3+b^3+c^3)\ge a^2b+ab^2+b^2c+bc^2+a^2c+ac^2
=ab(a+b)+bc(b+c)+ca(a+c)$
$=a^2(b+c)+b^2(c+a)+c^2(a+b)---(2)$
now we only have to prove (2) ,I will have a rest ,hope someone can finish it

 

[kaliprasad]

Spoiler

we have $a^3 + b^3 – a^2 b – b^2 a$
= $a^3 - a^2 b – b^2 a + b^3$
= $a^2(a-b)- b^2(a-b) = (a^2-b^2) (a-b) = (a+b)(a-b)^2> = 0$

Hence
$a^3 + b^3 > = a^2 b + b^2 a$

Multiply by 3 and add $a^3 + b^3$ on both sides

$4(a^3 + b^3) >= a^3 + b^3 + 3(a^2 b + b^2 a) > = (a+b)^3$
$4(a^3 + b^3) >= (a+b)^3$ .. (1)

Similarly

$4(b^3 + c^3) >= (b+c)^3$ ... (2)
$4(c^3 + a^3) >= (c+a)^3$ ...(3)

Adding (1), (2), (3) we get the result

 

[Albert1]

Spoiler

we have $a^3 + b^3 – a^2 b – b^2 a$
= $a^3 - a^2 b – b^2 a + b^3$
= $a^2(a-b)- b^2(a-b) = (a^2-b^2) (a-b) = (a+b)(a-b)^2> = 0$

Hence
$a^3 + b^3 > = a^2 b + b^2 a$

Multiply by 3 and add $a^3 + b^3$ on both sides

$4(a^3 + b^3) >= a^3 + b^3 + 3(a^2 b + b^2 a) > = (a+b)^3$
$4(a^3 + b^3) >= (a+b)^3$ .. (1)

Similarly

$4(b^3 + c^3) >= (b+c)^3$ ... (2)
$4(c^3 + a^3) >= (c+a)^3$ ...(3)

Adding (1), (2), (3) we get the result

$a^3 + b^3 \geq a^2 b + b^2a$---(i)
$b^3 + c^3 \geq b^2 c + c^2 b$---(ii)
$a^3 + c^3 \geq a^2 c + c^2 a$---(iii)
(i)+(ii)+(iii) is just the result of (2) of my previous post

 

[anemone]

Thanks to Albert and kaliprasad for participating and well done for cracking the problem in such a nice way!

Solution suggested by other:

Spoiler

We begin by computing

$\begin{align*} 4(a^3+b^3)-(a+b)^3&=(a+b)(4(a^2-ab+b^2)-(a+b)^2)\\&=(a+b)(3a^2-6ab+3b^2)\\&=3(a+b)(a-b)^2\\&\ge0 \end{align*}$

where the inequality holds since $a$ and $b$ are assumed to be positive, so $a+b>0$ and of course $(a-b)^2 \ge 0$.

Thus, $4a^3+4b^3>(a+b)^3$ and similarly, $4a^3+4c^3>(a+c)^3$ and $4b^3+4c^3>(b+c)^3$.

Adding these three inequalities we obtain

$8a^3+8b^3+8c^3>(a+b)^3+(a+c)^3+(b+c)^3$,

which is precisely what we wanted.