lehungvietbao wrote: Find all functions $\mathbb R^+\to\mathbb R^+$ such that \[(f(a)+f(b))(f(c)+f(d))=(a+b)(c+d), \quad \forall a,b,c,d\in\mathbb R^+; \quad abcd=1\] Let $P(x,y,z,t)$ be the assertion $(f(x)+f(y))(f(z)+f(t))=(x+y)(z+t)$, true $\forall x,y,z,t>0$ such that $xyzt=1$ $P(1,1,1,1)$ $\implies$ $f(1)=1$ $P(x,\frac 1x,x,\frac 1x)$ $\implies$ $f(\frac 1x)=x+\frac 1x-f(x)$ $P(x,1,\frac 1x,1)$ $\implies$ $(f(x)+1)(x+\frac 1x-f(x)+1)=(x+1)(\frac 1x+1)$ $\implies$ $(f(x)-x)(f(x)-\frac 1x)=0$ So $\forall x$, either $f(x)=x$, either $f(x)=\frac 1x$ Suppose now that $\exists u,v\ne 1$ such that $f(u)=u$ and $f(v)=\frac 1v$ We immediately get $f(\frac 1u)=\frac 1u$ and $f(\frac 1v)=v$ $P(u,v,\frac 1u,\frac 1v)$ $\implies$ $(u-1)(v-1)=0$, impossible. And so, either $\boxed{f(x)=x}$ $\forall x$, either $\boxed{f(x)=\frac 1x}$ $\forall x$, which indeed both are solutions.