Attention Required!

If $a,b,c$ và $d$ are positive real numbers satisfying the expression: $$a^4+b^4+c^4+d^4=4abcd$$ then, prove that $a=b=c=d$.

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$$a^4+b^4+c^4+d^4=4abcd$$$$a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4=4abcd-2a^2b^2-2c^2d^2$$$$(a^2-b^2)^2 +(c^2-d^2)^2 = -2(ab-cd)^2$$$$(a^2-b^2)^2 +(c^2-d^2)^2+2(ab-cd)^2=0$$

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 và 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 $d^4$, $a^2b^2$ and $c^2d^2$.

Xem thêm: Tạo Thiệp Chúc Mừng 20/11 Điện Tử, Thiệp Chúc Mừng Ngày Nhà Giáo Việt Nam 20/11

$$fraca^4 + b^42 ge sqrta^4 b^4 \fracc^4 + d^42 ge sqrtc^4 d^4 \fraca^2 b^2 + c^2 d^22 ge sqrta^2b^2c^2d^2$$

Summing up the first two và 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 and only if $a^4 = b^4$, $c^4 = d^4$ & $a^2 b^2 = c^2 d^2$ which means that $a = b = c =d$

tóm tắt
edited Apr 21, 2021 at 10:55
answered Apr 21, 2021 at 10:34

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