If `2x^3 +ax^2+ bx+ 4=0` (a and b are positive real numbers) has 3 real roots, then prove that `a+ b ge 6 (2^(1/3)+ 4^(1/3))`

Explanation

Let `alpha, beta, gamma` be the roots of `2x^(3) + ax^(2) + bx + 4 = 0`. Given that all the coefficients are positive, so all the roots will be negative.
Let `alpha_(1) = -alpha, alpha_(2) = -beta, alpha_(3) = - gamma " " `=>` alpha_(1) + alpha_(2) + alpha_(3) = (a)/(2)`
`alpha_(1) alpha_(2) + alpha_(2) alpha_(3) + alpha_(3) alpha_(1) = (b)/(2)`
`alpha_(1) alpha_(2) alpha_(3) = 2`
Applying `AM ge GM`, we have
`(alpha_(1) + alpha_(2) + alpha_(3))/(3) ge (alpha_(1) alpha_(2) alpha_(3))^(1//3) rArr a ge 6 xx 2^(1//3)`
Also `(alpha_(1) alpha_(2) + alpha_(2) alpha_(3) + alpha_(1) alpha_(3))/(3) gt (alpha_(1) alpha_(2) alpha_(3))^(2//3) " " rArr b ge 6 " " 4^(1//3)`
Therefore `a + b ge 6(2^(1//3) + 4^(1//3))`