hopeless404 wrote: Given any three real numbers $a,b,c,$ prove the equation has at least one real root. $$\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}-\frac{3}{x}=0$$ Wrong. Choose as counterexample $a=b=c=1$

$3(x+a)(x+b)(x+c)-x(x+a)(x+b)-x(x+b)(x+c)-x(x+a)(x+c)=0$ and $x(x+a)(x+b)(x+c) \neq 0$ Easy to check, that $0,-a,-b,-c$ are not solutions $c(x+a)(x+b)+a(x+b)(x+c)+b(x+a)(x+c)=0$ $(a+b+c)x^2+2(ab+bc+ca)x+3abc=0$ $D=4(ab+ac+bc)^2-12abc(a+b+c)=4( a^2b^2+b^2c^2+c^2a^2-abc(a+b+c))=2(a^2(b-c)^2+b^2(c-a)^2+c^2(a-b)) > 0$ so equation has two real roots.