Can You Prove Inequality Challenge II?

[anemone]

The cubic polynomial $x^3+mx^2+nx+k=0$ has three distinct real roots but the other polynomial $(x^2+x+2014)^3+m(x^2+x+2014)^2+n(x^2+x+2014)+k=0$ has no real roots. Show that $k+2014n+2014^2m+2014^3>\dfrac{1}{64}$.

 

[Opalg]

[sp]Let $p(x) = x^3+mx^2+nx+k$, and let $\alpha,\beta,\gamma$ be its three real roots, so that $p(x) = (x-\alpha)(x - \beta)(x - \gamma).$ Let $Y = 2014.$ If $p(x^2 + x + Y) = 0$ then $x^2+x+Y$ must be a root of $p(x)$, say $x^2+x+Y = \alpha.$ But if $p(x^2 + x + Y) = 0$ has no real roots then $x^2+x+ Y - \alpha = 0$ must have no real roots, so its discriminant must be negative: $1 - 4(Y- \alpha) < 0.$ Therefore $Y-\alpha > \frac14$, and similarly $Y-\beta > \frac14$, $Y-\gamma > \frac14$. It follows that $$ k+2014n+2014^2m+2014^3 = p(Y) = (Y -\alpha)(Y - \beta)(Y - \gamma) > \tfrac14\cdot \tfrac14\cdot \tfrac14 = \tfrac1{64}.$$[/sp]

 

[anemone]

[sp]Let $p(x) = x^3+mx^2+nx+k$, and let $\alpha,\beta,\gamma$ be its three real roots, so that $p(x) = (x-\alpha)(x - \beta)(x - \gamma).$ Let $Y = 2014.$ If $p(x^2 + x + Y) = 0$ then $x^2+x+Y$ must be a root of $p(x)$, say $x^2+x+Y = \alpha.$ But if $p(x^2 + x + Y) = 0$ has no real roots then $x^2+x+ Y - \alpha = 0$ must have no real roots, so its discriminant must be negative: $1 - 4(Y- \alpha) < 0.$ Therefore $Y-\alpha > \frac14$, and similarly $Y-\beta > \frac14$, $Y-\gamma > \frac14$. It follows that $$ k+2014n+2014^2m+2014^3 = p(Y) = (Y -\alpha)(Y - \beta)(Y - \gamma) > \tfrac14\cdot \tfrac14\cdot \tfrac14 = \tfrac1{64}.$$[/sp]

Bravo, Opalg!...and thanks for participating!:)