$a^4+b^4+c^4+d^4=4abcd$ , prove that $a=b=c=d$.

Question

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

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!

Answer 1

Using AM-GM with $a^4, b^4, c^4,d^4$, we get

$$ \frac{a^4 + b^4 + c^4 + d^4}{4} \ge \sqrt[4]{a^4 b^4 c^4 d^4} $$

This means that

$$ a^4 + b^4 + c^4 + d^4 \ge 4abcd $$

With equality if and 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 $b^4$, $c^4$ and $d^4$, $a^2b^2$ and $c^2d^2$.

$$ \frac{a^4 + b^4}{2} \ge \sqrt{a^4 b^4} \\ \frac{c^4 + d^4}{2} \ge \sqrt{c^4 d^4} \\ \frac{a^2 b^2 + c^2 d^2}{2} \ge \sqrt{a^2b^2c^2d^2} $$

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 and only if $a^4 = b^4$, $c^4 = d^4$ and $a^2 b^2 = c^2 d^2$ which means that $a = b = c =d$