<s>[INST] Show $a^3+b^3+c^3+d^3 \le 27$ Given $a, b, c, d$ are real numbers and $a^2+b^2+c^2+d^2 = 9$ show that
$$a^3+b^3+c^3+d^3\le27$$
I have tried to do the following:
$$(a^2+b^2+c^2+d^2)^{\frac{3}{2}} = 27 \\ \text{after simplifying} \\ (a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=27$$
By setting the following:
$$(a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=(a^3+b^3+c^3+d^3) \\ \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=0 \\
(a^2+b^2)(c^2+d^2)=-(a^2b^2+c^2d^2) \\ (a^3+b^3+c^3+d^3)+\frac{3}{2}(a^2b^2+-(a^2b^2+c^2d^2)+c^2d^2)=27 \\ (a^3+b^3+c^3+d^3)=27$$
and so we have shown the inequality to hold - does this proof work?
 [/INST] Sorry didn't really read your solution, but this might work:
Clearly $|a|,|b|,|c|,|d|\leq 3$ and thus $a^3\leq 3a^2$ and so on... and we are done.
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