## Ishaan94 3 years ago Show that if $$x^4+ax^3+2x^2+bx+1$$ has a real solution, then $$a^2+b^2 ≥ 8$$.

1. Ishaan94

@mukushla @KingGeorge

2. satellite73

taxing my brain first thing in the morning one thought: we know generally what this looks like. if it is to have at least one real solution, then at least one zero of the derivative must have second coordinate below the $$x$$ axis let me see if i can make that lead to anything

3. satellite73

seems that if either $$a=3$$ or $$b=3$$ you get a zero at $$x=-1$$

4. mukushla

here is my solution but i dont like it i will try to find a better one let $y=ax+\frac{b}{x}$ u have $x^4+ax^3+2x^2+bx+1=x^4+2x^2+1+x^2(ax+\frac{b}{x})=(x^2+1)^2+x^2y=0$ then $y=-(x+\frac{1}{x})^2$ there is at least one solution for equation above if and only if $y \le -4 \ \ \ \ since \ \ \ x+\frac{1}{x}\ge2$ then we must have $\max \left\{ y=ax+\frac{b}{x} \right\} =-4 \\ y'=0 \\ x=-\sqrt{\frac{b}{a}} \\ so \\ \max \left\{ y \right\}=-2\sqrt{ab}=-4 \\ ab=4 \\ a^2+b^2\ge 2ab=8$ a and be must have the same sign

5. mukushla

a better way to proof the second part we want to have $ax+\frac{b}{x}\le-4$ multiply it by x u will get $ax^2+4x+b \ge0 \\ or \\ ax^2+4x+b \le0$ since u dont know about the sign of x But In both cases discriminant of quadratic must be less than or equal to zero this gives: $16-4ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8$

6. mukushla

here is a better one $x^4 +ax^3+2x^2+bx+1 = (x^2 + \frac{a}{2} x)^2 + (1+\frac{ b}{2} x)^2 + \frac{1}{4} (8−a^2 −b^2)x^2$ In this case, the polynomial is strictly positive unless a^2+b^2−8 ≤ 0

7. Ishaan94

Perfect. I was trying to complete square but didn't get it. :/

8. Ishaan94

thanks a lot mukushla