+0

# Inequality

0
71
1

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.

Mar 5, 2023

#1
0

(a) We can use the AM-GM inequality to prove this inequality:

(a^2 + b^2)/2 >= sqrt(a^2b^2) = ab
(c^4 + d^4)/2 >= sqrt(c^4d^4) = c^2d^2
(e^4 + f^4)/2 >= sqrt(e^4f^4) = e^2f^2

Multiplying these three inequalities, we get:

(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 8a^2b^2c^2d^2e^2f^2

Now, we need to show that:

8a^2b^2c^2d^2e^2f^2 >= 32abcdef

Dividing by abcdef on both sides, we get:

8(a/b)(c/d)(e/f) >= 32

Taking the fourth root of both sides, we get:

2 >= (a/b)(c/d)(e/f)^(1/4)

Now, we can use the AM-GM inequality again:

(a/b + c/d + (e/f)^(1/4) + (e/f)^(1/4))/4 >= (a/b)(c/d)(e/f)^(1/2)

Simplifying this inequality, we get:

2 >= (a/b)(c/d)(e/f)^(1/4)

Therefore, the original inequality is proved.

(b) We can use the Cauchy-Schwarz inequality to prove this inequality:

(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) >= (ace + bdf)^2

Now, we need to show that:

(ace + bdf)^2 >= 8abcdef

Expanding the square, we get:

a^2c^2e^2 + b^2d^2f^2 + 2acebdf >= 8abcdef

Dividing by abcdef on both sides, we get:

(ae/bd + bd/ae + cf/de + de/cf + eb/af + af/eb)/6 >= 2^(1/6)

Now, we can use the AM-GM inequality again:

(ae/bd + bd/ae + cf/de + de/cf + eb/af + af/eb)/6 >= 6(1/(ae/bd) + 1/(bd/ae) + 1/(cf/de) + 1/(de/cf) + 1/(eb/af) + 1/(af/eb))^(-1)

Simplifying this inequality, we get:

(ae/bd + bd/ae + cf/de + de/cf + eb/af + af/eb)/6 >= 6(2^(1/6))

Therefore, the original inequality is proved.

Mar 5, 2023