Inequality

For part (a), we can use the AM-GM (Arithmetic Mean-Geometric Mean) inequality, which states that for any positive real numbers a_1, a_2, ..., a_n, their arithmetic mean is always greater than or equal to their geometric mean:

$(a_1 + a_2 + ... + a_n) / n >= sqrt[(a_1 * a_2 * ... * a_n)^(1/n)]$

Using this inequality repeatedly, we can prove the given inequality as follows:

$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) = [(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)]$

By AM-GM,

$[(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)] >= [(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]]$

And by AM-GM again,

$[(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]] >= [(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]]$

Finally, by AM-GM once more,

$[(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]] >= [(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2$

Applying AM-GM to the expression inside the square root on the right-hand side, we have

$[(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2 >= [(a^2 + b^2)^(1/2)][2(abcdef)^(1/6)]^2$

Finally, squaring both sides, we get

$[(a^2 + b^2)^2][(c * d * e * f)^(1/2)]^2 >= 32abcdef$, as desired.

For part (b), we can use the AM-GM inequality directly:

$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) = (a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)]$

By AM-GM,

$(a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)] >= (a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)]$

By AM-GM again,

$(a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)] >= 2(a^2$