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Question Number 33569 by Joel578 last updated on 19/Apr/18

$$\mathrm{Given}f\left(x\right)=x^3+{ax}^2+bx+c$$$$\mathrm{with}a,b,c\in \mathbb{R},\mathrm{the}\mathrm{roots}\mathrm{are}{x}_1,x_2,x_3\in \mathbb{R}$$$$\mathrm{Let}\lambda \mathrm{is}\mathrm{an}\mathrm{positive}\mathrm{integer}\mathrm{that}\mathrm{satisfied}$$$$x_2-x_1=\lambda$$$$x_3>\frac{1}{2}(x_1+x_2)$$$$\mathrm{What}\mathrm{is}\mathrm{the}\max \mathrm{value}\mathrm{of}\frac{2a^3+27c-9ab}{{\lambda }^3}?$$

Answered by MJS last updated on 20/Apr/18

$$f\left(x\right)=(x-x_1)(x-x_2)(x-x_3)=$$$$=x^3-(x_1+x_2+x_3)x^2+(x_1{x}_2+x_1{x}_3+x_2{x}_3)x-x_1{x}_2{x}_3$$$$a=-x_1-x_2-x_3$$$$b=x_1{x}_2+x_1{x}_3+x_2{x}_3$$$$c=-x_1{x}_2{x}_3$$$$$$$$x_2=x_1+\lambda$$$$a=-2x_1-x_3-\lambda$$$$b=x_1^2+2x_1{x}_3+\lambda (x_1+x_3)$$$$c=-x_1^2{x}_3-\lambda x_1{x}_3$$$$$$$$\frac{2a^3+27c-9ab}{{\lambda }^3}=$$$$=-3\frac{x_1-x_3}{\lambda }+3{\left(\frac{x_1-x_3}{\lambda }\right)}^2+2{\left(\frac{x_1-x_3}{\lambda }\right)}^3-2=t\left(\lambda \right)$$$$t^{\prime }\left(\lambda \right)=3\frac{x_1-x_3}{{\lambda }^2}-6\frac{{(x_1-x_3)}^2}{{\lambda }^3}-6\frac{{(x_1-x_3)}^3}{{\lambda }^4}$$$$t^{\prime }\left(\lambda \right)=0$$$$3(x_1-x_3){\lambda }^2-6{(x_1-x_3)}^2\lambda -6{(x_1-x_3)}^3=0$$$${\lambda }^2-2(x_1-x_3)\lambda -2{(x_1-x_3)}^2=0$$$$\lambda =(x_1-x_3)\pm \sqrt{3}(x_1-x_3)=(x_1-x_3)(1\pm \sqrt{3})$$$$$$$$x_3>\frac{1}{2}(x_1+x_2)\wedge x_2=x_1+\lambda \Rightarrow x_3>x_1+\frac{\lambda }{2}$$$$\lambda >0\wedge x_3>x_1+\frac{\lambda }{2}\Rightarrow \lambda =(x_1-x_3)(1-\sqrt{3})$$$$$$$$t\left(\lambda \right)=t\left((x_1-x_3)(1-\sqrt{3})\right)=\frac{3\sqrt{3}}{2}$$$$$$$$\max \left(\frac{2a^3+27c-9ab}{{\lambda }^3}\right)=\frac{3\sqrt{3}}{2}$$

Commented by Joel578 last updated on 20/Apr/18

$$thankyouverymuch$$

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