Raja Oktovin wrote: Consider a polynomial with coefficients of real numbers $ \phi(x) = ax^3 + bx^2 + cx + d$ with three positive real roots. Assume that $ \phi(0) < 0$, prove that \[ 2b^3 + 9a^2d - 7abc \le 0.\] Hery Susanto, Malang Let $ x_1,x_2$ and $ x_3$ be the roots of $ \phi(x)$,then : $ \phi(x) = a(x - x_1)(x - x_2)(x - x_3) = ax^3 - a(x_1 + x_2 + x_3)x^2 +$ $ a(x_1.x_2 + x_2.x_3 + x_3.x_1)x - a(x_1.x_2.x_3)$ Therefore : $ b = - a(x_1 + x_2 + x_3)$ , $ c = a(x_1.x_2 + x_2.x_3 + x_3.x_1)$ and $ d = - a(x_1.x_2.x_3)$, Since $ \phi(0) < 0$,If follows that $ a(x_1.x_2.x_3) > 0$ which lead to $ a > 0$, We have to prove that : $ 2(x_1 + x_2 + x_3)^3 + 9(x_1.x_2.x_3) \geq 7(x_1 + x_2 + x_3)(x_1.x_2 + x_2.x_3 + x_3.x_1)$ From Schur's inequality ( 3rd degree ) we have : $ x_1.x_2.x_3 \geq \frac { 4(x_1 + x_2 + x_3)(x_1.x_2 + x_2.x_3 + x_3.x_1) - (x_1 + x_2 + x_3)^3 }{9}$, Hence,It suffices to prove that : $ \left(x_1 + x_2 + x_3\right) \left( (x_1 + x_2 + x_3)^2 - 3(x_1.x_2 + x_2.x_3 + x_3.x_1) \right) \geq 0$, Which is true because : $ x_1 + x_2 + x_3 \geq 0$ and $ (x_1 + x_2 + x_3)^2 - 3(x_1.x_2 + x_2.x_3 + x_3.x_1)$ $ = \frac {(x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2}{2} \geq 0$

Assuming the roots of the equation are $ u,v,w$ and by viete's theorem $ b = - a(u + v + w) \ , \ c = a(uv + vw + wu) \ , \ d = - a(uvw)$ $ \phi(0) = d = - a(uvw) < 0$ , since $ u,v,w > 0$ , one can easily conclude that $ a > 0$ So it equivalent if we prove $ - 2a^3(u + v + w)^3 + 9a^2. - a(uvw)\le7a. - a(u + v + w).a(uv + vw + wu)$ $ 2(u + v + w)^3 + 9uvw\ge7(u + v + w)(yv + vw + wu)$ $ \left(2(u^3 + v^3 + w^3) + 6\sum u^2v + 12uvw\right) + 9uvw\ge7\sum u^2v + 21uvw$ $ 2(u^3 + v^3 + w^3)\ge\sum u^2v$ it is true since by AM-GM $ \sum (u^3 + u^3 + v^3) \ge 3\sum u^2v$ done

If $u,v,w$ are the roots then inequality is equivalent by Viete formulas to $2(u^3+w^3+v^3)\ge uv(u+v)+vw(v+w)+(u+w)uw$ which is true by AM GM

Let $r, s, t$ denote the roots of polynomial $P(x)$. By Vieta's Formulas: $$b = - a \sum r, c = a\sum rs, d = - a \sum rst.$$Since $P(x)$ has $3$ positive real roots and $P(0) < 0$, we must have: $P(x) \to -\infty$ as $x \to -\infty$ (else if $P(x) \to \infty$, then it would have a negative real root by Intermediate Value Theorem). Thus, $a>0$. Substituting the Vieta's formula into the expression: $S = 2b^3+9a^2d-7abc$ we have: $$S = a^3 \left( -2 (r+s+t)^3 - 9 rst + 7(r+s+t)(rs+st+tr) \right)$$Since $a > 0$, it suffice to show that: $$2 (r+s+t)^3+ 9 rst \ge 7(r+s+t)(rs+st+tr)$$inorder to have $S < 0$. Claim: $2(r^3+s^3+t^3) \ge \sum_{sym} r^2 s$ Proof: The above inequality is true due to Muirhead (or AM-GM). $$2(r^3+s^3+t^3) \ge \sum_{sym} r^2 s$$Adding $6 \sum_{sym} r^2 s+21rst$ to both sides of the above inequality, we have: $$2(\sum_{cyc} r^3 + 3 \sum_{sym} r^2s + 6 rst) + 9 rst \ge 7 (\sum_{sym} r^2 s + 3 rst)$$ $$\implies 2(r+s+t)^3+9rst \ge 7(r+s+t)(rs+st+tr).$$ Therefore the above inequality is true, which implies $S \le 0$ and we are done. Hence proved.