If $$2x^3+ax^2+bx+4=0$$ (a and b are positive real numbers) has $$3$$ real roots, then prove that $$ a+b \geq 6(2^{1/3}+4^{1/3})$$.
$$2x^3+9x^2+bx+4=0$$
Let the three real roots are $$\alpha,\beta,\gamma$$
Sum of roots, $$\alpha,\beta,\gamma=\dfrac{-a}{2}$$
$$\Rightarrow \alpha\beta+\beta\gamma+\gamma\alpha=\dfrac{b}{2}$$
$$\Rightarrow \alpha\beta\gamma=-2$$
If $$\alpha,\beta,\gamma$$ are three numbers then,
$$ AM\ge GM$$
$$\dfrac{\alpha+\beta+\gamma}{3}\ge \left(\alpha\beta\gamma\right)^{1/3}$$
$$\dfrac{-a}{2}\times \dfrac{1}{3}\ge \left(-2\right)^{1/3}$$
$$\dfrac{-a}{b}\ge -\left(2\right)^{1/3}\rightarrow \left(I\right)$$
$$\&$$ If $$\alpha\beta,\beta\gamma,\gamma\alpha$$ are three numbers then,
$$AM\ge GM$$
$$\dfrac{\alpha\beta+\beta\gamma+\gamma\alpha}{3}\ge \left(\alpha^2\beta^2\gamma^2\right)^{1/3}$$
$$\dfrac{b}{2}\times \dfrac{1}{3}\ge \left(-2^2\right)^{1/3}$$
$$\dfrac{b}{6}\ge -\left(4\right)^{1/3}\rightarrow \left(II\right)$$
From equation $$II$$ $$-$$ equation $$I$$
$$\dfrac{b}{6}+\dfrac{a}{b}\ge 4^{1/3}+2^{1/3}$$
$$a+b\ge 6\left(2^{1/3}+4^{1/3}\right)$$
Hence, the answer is $$a+b\ge 6\left(2^{1/3}+4^{1/3}\right).$$
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