Consider the equation $ x^4 - 4x^3 + 4x^2 + ax + b = 0$, where $ a,b\in\mathbb{R}$. Determine the largest value $ a + b$ can take, so that the given equation has two distinct positive roots $ x_1,x_2$ so that $ x_1 + x_2 = 2x_1x_2$.
Write $ P$ as $ P(x)=(x^2+Ax+B)(x^2-2t+t)$ where $ t=x_1x_2>0$
This gives:
$ A=2t-4;B=4t^2-9t+4;a=At-2Bt;b=Bt$
$ \implies a+b=(A-B)t=-4t^3+11t^2=f(t)$
$ f'(t)=-12t^2+22t=0$
$ \implies t=0;t=\frac{22}{12}$ but $ f(0)=0 ;f(\frac{22}{12})=\frac{484}{33};f(+\infty)=-\infty$
$ \implies Max(a+b)=\frac{484}{33}$
this Max is realised for $ x_1=\frac{22+\sqrt{220}}{12}>0;x_2=\frac{22-\sqrt{220}}{12}>0$
AYMANE wrote:
Write $ P$ as $ P(x)=(x^2+Ax+B)(x^2-2t+t)$ where $ t=x_1x_2>0$
This gives:
$ A=2t-4;B=4t^2-9t+4;a=At-2Bt;b=Bt$
$ \implies a+b=(A-B)t=-4t^3+11t^2=f(t)$
$ f'(t)=-12t^2+22t=0$
$ \implies t=0;t=\frac{22}{12}$ but $ f(0)=0 ;f(\frac{22}{12})=\frac{484}{33};f(+\infty)=-\infty$
$ \implies Max(a+b)=\frac{484}{33}$
this Max is realised for $ x_1=\frac{22+\sqrt{220}}{12}>0;x_2=\frac{22-\sqrt{220}}{12}>0$
Actually, this is wrong.
Let $x_1x_2=\lambda$, then $\lambda \geq 1$ by AM-GM. After some calculations, we have
$$a+b=-4\lambda^3+11\lambda^2-8\lambda=f(\lambda)$$Thus,
$$f^\prime(\lambda)=-2(2\lambda -1)(3\lambda -4)$$Because of $\lambda \geq 1$ and $\lim_{\lambda\to \infty}{f(\lambda)}=-\infty$, the maximum value of $a+b$ is $f(\tfrac{4}{3})$. $\blacksquare$