Question

$\sf If \: a^4 \: + \: b^4 \: + \: c^4 \: + \: d^4 \: = \: 4abcd$

$\sf Prove \: that \: !!$


$\sf \implies a \: = \: b \: = \: c \: = \: d$

Answer 1

Hi there!
Here's the answer:

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•

Given,
a⁴ + b⁴ + c⁴ + d⁴ = 4abcd
=> a⁴ + b⁴ + c⁴ + d⁴ - 4abcd = 0
--------------(1)

Add and subtract 2a²b² & 2c²d²

a⁴ + b⁴ + c⁴ + d⁴ - 2a²b² - 2c²d² +
2a²b² + 2c²d² - 4abcd = 0

Rearrange terms

=> (a⁴ + b⁴ - 2a²b²) + (c⁴ + d⁴ - 2c²d² ) - (2a²b² + 2c²d² - 4abcd) = 0

=> (a² - b²)² + (c² - d²)² +2(a²b² + c²d² - 2abcd) = 0

=> (a² - b²)² + (c² - d²)² +2(ab - cd)² = 0

---------------(2)

As when a term is squared, the No. is always positive.

a² - b² = 0 & c² - d² = 0 & ab - cd = 0

=> a² = b² & c² = d² & a/c = d/b

---------------(3)

Substitute in Eq.(1)

=> (a⁴ + c⁴ - 2a²c²) + (b⁴ + d⁴ - 2b²d² ) - (2a²c² + 2b²d² - 4abcd) = 0

=> (a⁴ + c⁴ - 2a²c²) + (b⁴ + d⁴ - 2b²d² ) - 2(a²c² + b²d² - abcd) = 0

=> (a² - c²)² + (b² - d²)² + 2(ac - bd)² = 0

Similar to earlier.

a² - c² = 0 & b² - d² = 0 & ac = bd

a² = c² & b² = d² & ac = bd

----------------(4)

From Eq(3) & Eq(4)

a = b = c = d

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•

Hope it helps

Answer 2

$\underline{\underline{\mathfrak{\large{solution:}}}}$

$\underline{\textsf{Given,}},\\\\ \sf\implies a^4\:+\:b^4\:+c^4\:+\:d^4\:=\:4abcd$

$\sf\implies\:a^4\:+b^4\:+c^4\:+d^4\:-4abcd\:=\:0\:\:\:....(1)$

$\sf\star\:Add \:and\: subtract \:2a^2b^2 \:and\: 2c^2d^2\:\star$

$\bf\implies\:a^4\:+b^4\:+c^4\:+d^4\:-2a^2b^2\:-2c^2d^2\:+2a^2b^2\:+2c^2d^2\:-4abcd\:=\:0$

$<b><u>Rearrange in terms:$

$\bf\sf\implies\:(a^2\:-b^2\:)\:=\:0$