# How to satisfy an inequality, from a fourth degree polynomial [duplicate]

If the equation $$P(x)=x^4+ax^3+2x^2+bx+1=0$$ has real solutions,
prove that $$a^2+b^2\geq8$$.

For the above question, I tried taking advantage of Vieta's formulas such as for the roots of the equation $$r_1, r_2 , r_3 , r_4$$, $$r_1+r_2+r_3+r_4=-a/1$$ and $$r_1r_2+r_1r_3+r_1r_4+...r_3r_4=2/1$$ etc. and afterwards I attempted to formulate a quadratic equation. Through this I attempted to get an inequality of the type $$b^2-4ac\geq0$$, so that I could get that $$a^2+b^2\geq8$$. However I did not succeed in doing this and hence solving the question. Can someone please explain to me how this question could be solved and if there is a better approach than the one I attempted to use?

• – Z Ahmed Jan 24 '20 at 10:31

$$P(x) = x^4+ax^3+2x^2+bx+1 = x^2 \left(x + \frac a2 \right)^2 + \left(\frac b2 x + 1 \right)^2 + \left( 2 - \frac{a^2+b^2}{4}\right) x^2$$ has no real zeros if $$a^2+b^2 < 8$$.
The bound is sharp: For $$a=b=2$$ we have $$a^2+b^2=8$$ and $$P(x) = x^4+2x^3+2x^2+2x+1 = (x+1)^2(x^2+1)$$ with a real zero at $$x=-1$$.
By Cauchy's inequality, $$(a^{2} + b^{2})(x^{6} + x^{2}) \geq (ax^{3} + bx)^{2} = (-(x^{4}+2x^{2}+1))^{2} = (x^{2}+1)^{4}$$ so $$a^{2} +b^{2} \geq \frac{(x^{2} + 1)^{4}}{x^{6} + x^{2}}$$ Now try to show that the RHS attains minimum value $$8$$ when $$x = 1$$. One way is to use substitution $$t=x + \frac{1}{x}$$ and then $$\frac{(x^{2} + 1)^{4}}{x^{6} + x^{2}} = \frac{t^{4}}{t^{2} - 2} \geq 8 \Leftrightarrow (t^{2} - 4)^{2} \geq 0$$