Bạn đang xem: Algebra precalculus
$$a^4+b^4+c^4+d^4=4abcd$$$$a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4=4abcd-2a^2b^2-2c^2d^2$$$$(a^2-b^2)^2 +(c^2-d^2)^2 = -2(ab-cd)^2$$$$(a^2-b^2)^2 +(c^2-d^2)^2+2(ab-cd)^2=0$$
so then we get
$a^2=b^2;$ $c^2=d^2;$ and $ab=cd$.
I am looking for a method using $AM-GM$ inequality. Any help would be most appreciated!
Using AM-GM with $a^4, b^4, c^4,d^4$, we get
$$fraca^4 + b^4 + c^4 + d^44 ge sqrt<4>a^4 b^4 c^4 d^4$$
This means that
$$a^4 + b^4 + c^4 + d^4 ge 4abcd$$
With equality if & only if $a^4 = b^4 = c^4 = d^4 implies a = b = c =d$
Alternatively, we can just transform your proof into an AM-GM one:
We use AM-GM on $a^4$ and $b^4$, $c^4$ và $d^4$, $a^2b^2$ & $c^2d^2$.
$$fraca^4 + b^42 ge sqrta^4 b^4 \fracc^4 + d^42 ge sqrtc^4 d^4 \fraca^2 b^2 + c^2 d^22 ge sqrta^2b^2c^2d^2$$
Summing up the first two and then using the third inequality we get,$$a^4 + b^4 + c^4 + d^4 ge 2(a^2 b^2 + c^2 d^2) ge 2 cdot 2abcd ge 4abcd$$
With equality if và only if $a^4 = b^4$, $c^4 = d^4$ và $a^2 b^2 = c^2 d^2$ which means that $a = b = c =d$
edited Apr 21, 2021 at 10:55
answered Apr 21, 2021 at 10:34
63033 silver badges1414 bronze badges
địa chỉ cửa hàng a phản hồi |
Thanks for contributing an answer lớn onfire-bg.comematics Stack Exchange!Please be sure khổng lồ answer the question. Provide details và share your research!
But avoid …Asking for help, clarification, or responding lớn other answers.Making statements based on opinion; back them up with references or personal experience.
Use onfire-bg.comJax khổng lồ format equations. onfire-bg.comJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
Sign up using Google
Sign up using Facebook
Sign up using e-mail and Password
Post as a guest
email Required, but never shown
Post as a guest
thư điện tử
Required, but never shown
Post Your Answer Discard
Featured on Meta
Prove that $frac4abcd geq frac a b + frac bc + frac cd +frac d a$
Contest Inequality - Is it AM GM?
Prove: $frac1x-a+frac1x-b+frac1x=0$ has a real root between $frac13a$ & $frac23a$, & one between $-frac23b$ & $-frac13b$
For positive real numbers $a,b,c$ prove that $2+fraca^6+b^6+c^63ge ab+ac+bc$
Prove that $displaystyleprod_k=1^n frac1+x_kx_k geq prod_k=1^n fracn-x_k1-x_k$
Hot Network Questions more hot questions
Subscribe to RSS
Question feed lớn subscribe to this RSS feed, copy and paste this URL into your RSS reader.
Stack Exchange Network
Site thiết kế / logo sản phẩm © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.6.17.42396