homeschool girl
Junior Member
- Joined
- Feb 6, 2020
- Messages
- 123
The Problem:
Let a, b, c, d, e, f be nonnegative real numbers.
(a) Prove that
[math](a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) \ge (ace + bdf)^4.[/math](b) Prove that
[math](a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.[/math]
Hints:
(a) Work with [imath](c^4 + d^4)(e^4 + f^4)[/imath] first.
(b) Try proving an inequality of the form [imath](a^2 + b^2)(\text{expression}) \ge (ace + bdf)^2[/imath].
Where I'm at so far:
for part a, I used the Cauchy-Schwarz inequality on [imath](a, b),[/imath] and [imath](ce, df)[/imath] to get
[imath](c^4 + d^4)(e^4 + f^4) \ge (c^2e^2 + d^2f^2)^2[/imath]
then I multiplied both sides by [imath](a^2+b^2)^2[/imath]
[imath](a^2+b^2)^2(c^4 + d^4)(e^4 + f^4) \ge (a^2+b^2)^2(c^2e^2 + d^2f^2)^2[/imath]
the expansion of [imath](ace + bdf)^4[/imath] is is close to the expansion of [imath](a^2+b^2)^2(c^2e^2 + d^2f^2)^2[/imath], but not quite, and I'm not sure how I could prove [imath](a^2+b^2)^2(c^2e^2 + d^2f^2)^2 \ge (ace + bdf)^4[/imath].
for part b, I used the Cauchy-Schwarz inequality on [imath](c^2, d^2),[/imath] and [imath](e^2, f^2)[/imath] :
[imath](a^2 + b^2)(c^2e^2+d^2f^2) \ge (ace + bdf)^2,[/imath]
which is close to the given inequality, but [imath](c^2 + d^2)(e^2 + f^2) = c^2e^2+d^2f^2+c^2f^2+d^2e^2[/imath], not [imath]c^2e^2+d^2f^2[/imath].
however, if i could prove that [imath](a^2 + b^2)(c^2f^2+d^2e^2) \ge 0[/imath], then i could subtract it from [imath](a^2 + b^2)(c^2e^2+d^2f^2) \ge (ace + bdf)^2[/imath], to get
[imath](a^2 + b^2)(c^2e^2+d^2f^2+c^2f^2+d^2e^2)-(a^2 + b^2)(c^2f^2+d^2e^2)\ge (ace + bdf)^2[/imath], which simplifies to
[imath](a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.[/imath]
but i don't know how to go about proving [imath](a^2 + b^2)(c^2f^2+d^2e^2) \ge 0[/imath].
I'm completely stuck on how to finish the problem and at this point, I don't even know if I'm on the right track, so any help is appreciated, Thanks!
Let a, b, c, d, e, f be nonnegative real numbers.
(a) Prove that
[math](a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) \ge (ace + bdf)^4.[/math](b) Prove that
[math](a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.[/math]
Hints:
(a) Work with [imath](c^4 + d^4)(e^4 + f^4)[/imath] first.
(b) Try proving an inequality of the form [imath](a^2 + b^2)(\text{expression}) \ge (ace + bdf)^2[/imath].
Where I'm at so far:
for part a, I used the Cauchy-Schwarz inequality on [imath](a, b),[/imath] and [imath](ce, df)[/imath] to get
[imath](c^4 + d^4)(e^4 + f^4) \ge (c^2e^2 + d^2f^2)^2[/imath]
then I multiplied both sides by [imath](a^2+b^2)^2[/imath]
[imath](a^2+b^2)^2(c^4 + d^4)(e^4 + f^4) \ge (a^2+b^2)^2(c^2e^2 + d^2f^2)^2[/imath]
the expansion of [imath](ace + bdf)^4[/imath] is is close to the expansion of [imath](a^2+b^2)^2(c^2e^2 + d^2f^2)^2[/imath], but not quite, and I'm not sure how I could prove [imath](a^2+b^2)^2(c^2e^2 + d^2f^2)^2 \ge (ace + bdf)^4[/imath].
for part b, I used the Cauchy-Schwarz inequality on [imath](c^2, d^2),[/imath] and [imath](e^2, f^2)[/imath] :
[imath](a^2 + b^2)(c^2e^2+d^2f^2) \ge (ace + bdf)^2,[/imath]
which is close to the given inequality, but [imath](c^2 + d^2)(e^2 + f^2) = c^2e^2+d^2f^2+c^2f^2+d^2e^2[/imath], not [imath]c^2e^2+d^2f^2[/imath].
however, if i could prove that [imath](a^2 + b^2)(c^2f^2+d^2e^2) \ge 0[/imath], then i could subtract it from [imath](a^2 + b^2)(c^2e^2+d^2f^2) \ge (ace + bdf)^2[/imath], to get
[imath](a^2 + b^2)(c^2e^2+d^2f^2+c^2f^2+d^2e^2)-(a^2 + b^2)(c^2f^2+d^2e^2)\ge (ace + bdf)^2[/imath], which simplifies to
[imath](a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.[/imath]
but i don't know how to go about proving [imath](a^2 + b^2)(c^2f^2+d^2e^2) \ge 0[/imath].
I'm completely stuck on how to finish the problem and at this point, I don't even know if I'm on the right track, so any help is appreciated, Thanks!
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