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^(2) + 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`. Then,
`alpha_(1) + alpha_(2 + alpha_(3) = a//2`
`alpha_(1) alpha_(2) + alpha_(2)alpha_(3 + alpha_(3) alpha_(1) = b//2`
and `alpha_(1) alpha_(2) alpha_(3) = 2`
Now, `A.M ge G.M`
`implies (alpha_(1) + alpha_(2) + alpha_(3))/(3) ge (alpha_(2) alpha_(2) alpha_(3))^(1//3)`
or `(a^(3))/(216) ge 2`
or `a ge 6 xx 2^(1//3)`
Also, `(alpha_(2) alpha_(2) + alpha_(2)alpha_(3) + alpha_(1) alpha_(3))/(3) ge (alpha_(1) alpha_(2) alpha_(3))^(2//3)`
or `(b^(3))/(216) ge 4`
or `b ge 6 xx 4^(1//3)`
Adding Eqs. (1) and (2), we get
`a + b ge 6 (2^(1//3) + 4^(1//3))`