a_507_bc wrote: The equation $$t^4+at^3+bt^2=(a+b)(2t-1)$$has $4$ positive real roots $t_1<t_2<t_3<t_4$. Show that $t_1t_4>t_2t_3$. Let $r_i=\frac 1{t_i}$ roots of $P(x)=x^4-2x^3+\frac b{a+b}x^2+\frac a{a+b}x+\frac 1{a+b}$ Note that $P(1-x)=P(x)$ so that $r_1+r_4=r_2+r_3=1$ and, since $r_1>r_2>r_3>r_4>0$ and $r_1+r_2+r_3+r_4=2$ : $r_1+r_2>1$ So (multiplying by $r_1-r_2$) : $r_1^2-r_2^2>r_1-r_2$, which is $r_1(1-r_1)<r_2(1-r_2)$ and so $r_1r_4<r_2r_3$ And so $t_1t_4>t_2t_3$ Q.E.D

pco wrote: a_507_bc wrote: The equation $$t^4+at^3+bt^2=(a+b)(2t-1)$$has $4$ positive real roots $t_1<t_2<t_3<t_4$. Show that $t_1t_4>t_2t_3$. Let $r_i=\frac 1{t_i}$ roots of $P(x)=x^4-2x^3+\frac b{a+b}x^2+\frac a{a+b}x+\frac 1{a+b}$ Note that $P(1-x)=P(x)$ so that $r_1+r_4=r_2+r_3=1$ and, since $r_1>r_2>r_3>r_4>0$ and $r_1+r_2+r_3+r_4=2$ : $r_1+r_2>1$ So (multiplying by $r_1-r_2$) : $r_1^2-r_2^2>r_1-r_2$, which is $r_1(1-r_1)<r_2(1-r_2)$ and so $r_1r_4<r_2r_3$ And so $t_1t_4>t_2t_3$ Q.E.D Why $r_1+r_4=r_2+r_3$ ?

Jjesus wrote: Why $r_1+r_4=r_2+r_3$ ? Because $P(1-x)=P(x)$ and so $P(x)$ is symetric versus $x=\frac 12$