It is just AM-GM: $a+(b+4)\ge a+4\sqrt{b}\ge 4\sqrt{a\sqrt{b}}$ so $b=4 and a=8$.

parmenides51 wrote: Solve the equation $a + b + 4 = 4\sqrt{a\sqrt{b}}$ in real numbers We need $a,b\ge 0$ and equation is $(\sqrt a-2\sqrt[4]b)^2+(\sqrt b-2)^2=0$ And so $\boxed{(a,b)=(8,4)}$

$a=-4, b=0$ works, too...

socrates wrote: $a=-4, b=0$ works, too... Yea, maybe we should consider the case where b is equal to 0 separately because then, a doesn't necessarily have to be greater or equal to 0.

Solve for positive real numbers:

Attachments:

AM-QM $$\left(\frac{x+y}{2}\right)^5\le\frac{x^5+y^5}{2},\ \left(\frac{1+y}{2}\right)^5\le\frac{1+y^5}{2},\ \left(\frac{x+1}{2}\right)^5\le\frac{x^5+1}{2}$$Adding gives inequality with equality case $x=y=1$. AM-QM again $$\sqrt{xz}+\sqrt{\frac{x^2+z^2}{2}}\le 2\sqrt{\frac{xz+\frac{x^2+z^2}{2}}{2}}=x+z$$Equality iff $z=x.$ Answer: $$x=y=z=1.\blacksquare$$