$x_1+x_2+x_3=a>0,x_2x_3+x_3x_1+x_1x_2=b>0,x_1x_2x_3=a>0,$ $\frac{a}{3}=\frac{x_1+x_2+x_3}{3}\ge \sqrt[3]{x_1x_2x_3}=\sqrt[3]{a} \Rightarrow a\ge 3\sqrt{3},$ $ a^2=(x_1+x_2+x_3)^2\ge 3(x_2x_3+x_3x_1+x_1x_2) =3b\Rightarrow \frac{a^2+3}{b+1}\ge 3$, $\Rightarrow \frac{2a^3-3ab+3a}{b+1}=2a\cdot \frac{a^2+3}{b+1}-3a \ge 3a \ge 9\sqrt{3}$, $\left(\frac{2a^3-3ab+3a}{b+1}\right)_{min}=9\sqrt{3}.$

2011 Thailand Q7: Given positive real number $a,b,c,d$ if all roots of equation $x^5-ax^4+bx^3-cx^2+dx-1=0$ are real .Prove that\[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d} \leq \frac{3}{5}.\] 2002 China Second Round Olympiad Let $a,b,c$ be real numbers such that the equation $x^3+ax^2+bx+c=0$ has three positive real roots $x_1,x_2,x_3$ , and $x_1-x_2=\lambda>0 $ , $x_1>\frac{x_2+x_3}{2}$. Find the maximum of $\frac{2a^3+27c-9ab}{\lambda^3}$. 2012 China Baidu Math competition: Let $a,b$ be real numbers such that the equation $ax^3-x^2+bx-1=0$ has three positive real roots . Find the minimum of $\frac{5a^2-6ab+3}{a^3(b-a)}$. Let $x,y,z$ be positive numbers satisfying $x^2+y^2+z^2=1$ . Find the minimum of \[{\frac{x^5}{y^2+z^2-yz}+\frac{y^5}{z^2+x^2-zx}}+\frac{z^5}{x^2+y^2-xy}\].

Let $\alpha,\beta,\gamma$ be the roots.Then we have $\alpha+\beta+\gamma=a$ $\alpha\beta+\beta\gamma+\gamma\alpha=b$ $\alpha\beta\gamma=a$ Note that $\frac{a}{3}=\frac{\alpha+\beta+\gamma}{3} \ge (\alpha\beta\gamma)^{\frac{1}{3}}=a^{\frac{1}{3}} \implies a \ge 3\sqrt{3}$ $a^2=(\alpha+\beta+\gamma)^2 \ge 3(\alpha\beta+\beta\gamma+\gamma\alpha)=3b \implies a^3 \ge 3ab$ Thus $\frac{2a^3-3ab+3a}{b+1} \ge \frac{6ab-3ab+3a}{b+1}=\frac{3a(b+1)}{b+1}=3a \ge 9\sqrt{3}$ Equality holds when $a=3\sqrt{3},b=9$.

let $f(x)=x^3-ax^2+bx-a$ since roots of $f(x)$ are real and positive thus this implies that roots of $f'(x)$ must be also real. we get $f'(x) = 3x^2-2ax+b$ and its $D=4a^2-12b\ge 0$ which implies $a^2\ge 3b$ let roots of $f(x)=0$ be $p,q,r$ now by $AM-GM$ $(p+q+r)^3\ge 27pqr$ or $a^3\ge 27a$ or $a\ge 3\sqrt 3$ now again $ a^2\ge 3b$ implies $2a^3\ge 6ab$ or $2a^3-3ab+3a\ge 3ab+3a$ and hence $\frac{2a^3-3ab+3a}{b+1}\ge \frac{3ab+3a}{b+1} = 3a \ge 9\sqrt 3 $ and hence proved.