Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).

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Show that if \(x^4+ax^3+2x^2+bx+1\) has a real solution, then \(a^2+b^2 ≥ 8\).

Mathematics
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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
seems that if either \(a=3\) or \(b=3\) you get a zero at \(x=-1\)

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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
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: \[16-4ab \le0\\ab \ge4\\a^2+b^2\ge2ab \ge8\]
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
Perfect. I was trying to complete square but didn't get it. :/
thanks a lot mukushla

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