# 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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Approach:

\$\$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\$.

Hence:\$a=b=c=d\$.

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\$.

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\$\$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\$

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edited Apr 21, 2021 at 10:55
answered Apr 21, 2021 at 10:34 OussemaOussema
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