Prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$ if equation $2x^3+ax^2+bx+4=0$ has $3$ real roots

Peoblem Staement:-

If equation $2x^3+ax^2+bx+4=0$ has $3$ real roots ($a,b\gt0$), then prove that $a+b\ge6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$.

Attempt at a solution:-

Let the roots of the equation $2x^3+ax^2+bx+4=0$ be $\alpha_1, \alpha_2$ and $\alpha_3$.

We have, $$\sum_{i=1}^3{\alpha_i}=-\dfrac{a}{2}\\ \sum_{1\le i\lt j\le3}{\alpha_i\alpha_j}=\dfrac{b}{2}\\ \alpha_1\alpha_2\alpha_3=-\dfrac{4}{2}=-2$$

Now from AM-GM inequality we get, $$\dfrac{\displaystyle\sum_{i=1}^3{\alpha_i}}{3}\ge\sqrt{\prod_{i=1}^3{\alpha_i}}\implies -\dfrac{a}{6}\ge\sqrt{-2}\implies a\ge6\cdot{2}^{\frac{1}{3}}$$ Equality occurring iff $\alpha_1=\alpha_2=\alpha_3$

Similarly form the AM-GM inequality, we get $$\dfrac{\displaystyle\sum_{1\le i\lt j\le3}{\alpha_i\alpha_j}}{3}\ge \sqrt{\prod_{i=1}^3{\alpha_i}^2}\implies \dfrac{b}{6}\ge 4^{\frac{1}{3}}\implies b\ge 6\cdot{4}^{\frac{1}{3}}$$

Equality only occurring when $\alpha_1\alpha_2=\alpha_2\alpha_3=\alpha_3\alpha_1\implies\alpha_1=\alpha_2=\alpha_3$

Hence, $a+b\ge 6\cdot(2^{\frac{1}{3}}+4^{\frac{1}{3}})$

The solution looks satisfying but the only problem is that AM-GM inequality can be used only for non-negative numbers whereas $\sum_{i=1}^3{\alpha_i}$ is clearly a negative number as $a\gt0$. I was also not so sure about adding the two inequalities to get at the inequality the question asks to prove.

I am also not so sure when I can do this, i.e. add the inequalities. In this case is it valid due to the equality occurring at the same values in both the inequalities.

And, finally if you have a better solution please do post it.

• Your title statement does not agree with your peoblem statement. – Arthur Nov 16 '16 at 9:28
• @Arthur- I apologise, I had put the title of the question I was going to post but decided not to. – user350331 Nov 16 '16 at 9:36
• if a and b > 0, all real roots are less then 0 – kotomord Nov 16 '16 at 9:58
• @kotomord - Yeah that's what's bugging me if I could apply am gm inequality if all the roots are negative. – user350331 Nov 16 '16 at 10:01
• in (b) equation - root1*root2 >0 - you can yse AM-GM in(a) equation - work with (-root) – kotomord Nov 16 '16 at 10:09

Since $a, b > 0$, the polynomial is strictly positive for any nonnegative $x$. But this means that all $\alpha_i$ are negative. Now for example
$$2^{1/3} = (|\alpha_1| |\alpha_2| |\alpha_3|)^{1/3} \le \frac{1}{3}(|\alpha_1| + |\alpha_2| + |\alpha_3|) = -\frac{1}{3}(\alpha_1 + \alpha_2 + \alpha_3) = \frac{a}{2}.$$
Note that you also made a mistake when handling the inequalities, since $-\frac{a}{6} \ge -2^{1/3}$ is equivalent to $a \le 6 \cdot 2^{1/3}$ (multiplying an inequality with a negative number changes the direction of the inequality).
Adding both inequalities is also correct, since $a \ge b$ and $c \ge d$ implies $a + c \ge b + d$.
• I know that multiplying an inequality by a negative number changes the sign but here both are positive i.e. $a$ & $2$, hence the inequality is regarding their magnitudes in a way, isn't it? – user350331 Nov 16 '16 at 10:21
• But $-6$ is not a positive number and you need to multiply your inequality with $-6$ to get to $a \le 6\cdot 2^{1/3}$. – Dominik Nov 16 '16 at 10:24