Present Mystery

According to Cauchy-Schwarz inequality , $(a^2 + b^2 + c^2)(b^2 + c^2 + a^2) \geq (ab + bc + ca)^2$ .

Hence, $a^2 + b^2 + c^2 \geq ab + bc + ca$ .

Then, $6a^2 + 6b^2 + 6c^2 \geq 3(2ab + 2bc + 2ca)$ .

The total surface area of all pink cubes = $6a^2 + 6b^2 + 6c^2$ , and the total surface area of all blue cuboids = $3(2ab + 2bc + 2ca)$ .

Since $a, b, c$ are co-prime integers, the ratios $a:b \neq b:c \neq c:a$ .

As a result, such inequality can not become equality, and the pink cubes will always have larger total surface area than the blue ones.

The problem can be written as comparing $6a^2+6b^2+6c^2$ (LHS) and $6ab+6bc+6ca$ (RHS). Dividing 3 out and subtracting LHS-RHS, the remaining expression is $2a^2+2b^2+2c^2-2ab-2bc-2ca$ , which can be expressed as $(a-b)^2+(b-c)^2+(c-a)^2$ , obviously ≥0. Therefore, the LHS is greater, so pink has the bigger surface area. Equality occurs when a=b=c

According to AM -GM inequality, a^3 + b^3+ c^3 ≥ 3 ∛a^3b^3c^3 ⇒ a^3 + b^3+ c^3 ≥ 3 abc ⇒Sum of volumes of pink boxes ≥ Sum of volumes of blue boxes

Surface area of the three cubes= $6a^{2} + 6b^{2} + 6c^{2}$

surface area of the three cuboids= 3(2ab +2bc+ 2ac)= 6(ab+bc+ac)

Now according to cauchy shwarz inequality:

$(a^{2} + b^{2} + c^{2})( 1^{2} + 1^{2} +1^{2}) >= [a(1) +b(1) +c(1)]^{2}$

$3(a^{2} +b^{2} +c^{2}) >= (a+b+c)^{2}$

$3a^{2} +3b^{2} +3c^{2} >= a^{2} +b^{2} +c^{2} +2(ab+bc+ac)$

$2( a^{2} +b^{2} +c^{2}) >= 2(ab+bc+ac)$

multiplying 3 on both sides:

$6(a^{2} +b^{2} +c^{2}) >= 6(ab+bc+ac)$

therefore total surface area of the three cubes is greater than the total surface area of the three cuboids.

As $AM ≥ GM$ and $a ≠ b ≠ c$ from the fig.,

$a^2 + b^2 > 2ab$

$b^2 + c^2 > 2bc$

$a^2 + c^2 > 2ac$

Adding them up,

$\implies 2(a^2 + b^2 + c^2) > 2ab + 2bc + 2ac$

$\implies 6a^2 + 6b^2 + 6c^2 > 3(2ab + 2bc + 2ac)$

$\therefore \textcolor{#FFFFFF}{aaaa:} S.A.($ Cubes $)>S.A.($ Cuboids $)$

By the AM-GM inequality $a^{3}+b^{3}+c^{3}>3abc$

Rearrangement inequality.