Plz help prove this inequality

(a) Show that if a, b, c, d, e, f are nonnegative real numbers, then (a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 32abcdef.

(b) Show that if a, b, c, d, e, f are nonnegative real numbers, then (a^2 + b^2)(c^2 + d^2)(e^2 + f^2) >= 8abcdef.

Guest Feb 11, 2023

#1**0 **

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,

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

Mathefreaker2021 Feb 11, 2023