Question

Cho a^4+b^4+c^4+d^4=4abcd với a,b,c,d lá số thực dương.CMR: a=b=c=d

Answer 1

`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(a^2b^2-2abcd+c^2d^2)>=0`

`<=>(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2=0`

Vì `VT>=0AA a,b,c,d`

Dấu "=" xảy ra khi `a^2=b^2,c^2=d^2,ab=cd`

`<=>a=b=c=d`

Answer 2

áp dụng BDT AM-GM

\(=>a^4+b^4\ge2\sqrt{\left(ab\right)^4}=2a^2b^2\left(1\right)\)

\(=>c^4+d^4\ge2\sqrt{c^4d^4}=2c^2d^2\left(2\right)\)

(1)(2)\(=>a^4+b^4+c^4+d^4\ge2\left(a^2b^2+c^2d^2\right)\ge4abcd\)

dấu"=" xảy ra\(< =>\left\{{}\begin{matrix}a^4=b^4\\c^4=d^4\end{matrix}\right.< =>a=b=c=d}\)