Maximum of a Big Fraction

We know that $a^3 + b^3 = (a+b)(a^2 + b^2 -ab)$ . So, $\begin{aligned} & (a+b+c)^3 - a^3 - b^3 - c^3 \\ = & (b+c)(a^2+b^2+c^2+2ab+2bc+2ac + a^2 +a^2+ab+ac) -(b+c) (b^2 - c^2 +bc)\\ = & (b+c)(3a^2 + 3ab + 3bc + 3ac) \\ = & 3(a+b)(b+c)(a+c)\\ \end{aligned}$ Then the expression in the equation becomes $3\times \frac{(a+b)^2(b+c)^2(a+c)^2}{(a^2+b^2)(b^2+c^2)(a^2+c^2)}$ . Now we note that $\frac {(a+b)^2}{(a^2+b^2)} \leq 2$ , which shows that the expression is less than or equal to $3\times 2\times 2\times 2 = 24$ .

For $a_1, a_2, a_3..., a_n$ ,

We have $\frac{a_1+a_2+a_3...+a_n}{n}\leq\sqrt{\frac{a_1^2+a_2^2+a_3^2...+a_n^2}{n}}$ , By AM-QM

Multiplying by n, we get: $a_1+a_2+a_3...+a_n \leq \sqrt{(a_1^2+a_2^2+a_3^2...+a_n^2) \times n}$ ,

Squaring, we have: $(a_1+a_2+a_3...+a_n)^2 \leq (a_1^2+a_2^2+a_3^2...+a_n^2) \times n$ ,

Thus $\frac{(a_1+a_2+a_3...+a_n)^2}{(a_1^2+a_2^2+a_3^2...+a_n^2)} \leq n$ , true for $a_1=a_2=a_3...=a_n$

Let

$a_1=a_2=a_3=x^2y$ ,

$a_4=a_5=a_6=x^2z$ ,

$a_7=a_8=a_9=y^2x$ ,

$a_{10}=a_{11}=a_{12}=y^2z$ ,

$a_{13}=a_{14}=a_{15}=z^2z$ ,

$a_{16}=a_{17}=a_{18}=z^2z$ ,

$a_{19}=a_{20}=a_{21}=a_{22}=a_{23}=a_{24}=xyz$ ,

By evaluating, we see

$((x+y+z)^3-x^3-y^3-z^3)^2=(a_1+a_2+a_3...+a_{24})^2$

$((x^2+y^2+z^2)^3-x^6-y^6-z^6)^2=a_1^2+a_2^2+a_3^2...+a_{24}^2$

Thus $\frac{((x+y+z)^3-x^3-y^3-z^3)^2}{((x^2+y^2+z^2)^3-x^6-y^6-z^6)^2}= \frac{(a_1+a_2+a_3...+a_{24})^2}{(a_1^2+a_2^2+a_3^2...+a_{24}^2)} \leq 24$

We shall now show this is possible. Setting x=y=z=1,

$\frac{((x+y+z)^3-x^3-y^3-z^3)^2}{((x^2+y^2+z^2)^3-x^6-y^6-z^6)^2}=\frac{((1+1+1)^3-1^3-1^3-1^3)^2}{((1^2+1^2+1^2)^3-1^6-1^6-1^6)^2}=\frac{24^2}{24}=24$

Every term in $(x+y+z)^3-x^3 -y^3-z^3$ has its square in $(x^2+y^2+z^2)^3-x^6 -y^6-z^6$ , so by Cauchy–Schwarz inequality we get $((x^2+y^2+z^2)^3-x^6 -y^6-z^6)(24) \geq ((x+y+z)^3-x^3 -y^3-z^3)^2$ , with equality reached if and only if $x=y=z$ .

From big's theoreme number 1 x=y=z=1 so the answer is 24²:24 = 24

:V

Assuming the value will be maximized when x=y=z, we can substitute x for y and z and create the new fraction $\frac{((3x)^{3}-3x^{3})^{2}}{(3x^{2})^{3}-3x^{6}}$ which simplifies to $\frac{(24x^{3})^{2}}{24x^{6}}$ or $\frac{24^{2}x^{6}}{24x^{6}}$ which finally simplifies to 24.

The maximum possible values of the expression will be attained when all the three variables x,y and z are equal. In that case the expression will become, {(3x)^3 -3x^3}^2 /(3x^2 )^3 -3 x^6 = 24

Use holders inequality