Solving Equation through inequalities.
If $x^6-12x^5+ax^4+bx^3+cx^2+dx+64=0$ has positive roots then find $a,b,c,d$.
I did something but that don't deserve to be added here, but what I thought before doing that is following:
- For us, Product and Sum of roots are given.
- Roots are positive.
- Hence I tried to use AM-GM-HM inequalities, as sum and product are given, but I failed to conclude something good.
So please deliver some hints or solution regarding AM-GM-HM inequalities.
Use AM-GM on the roots.
Say $a_1,a_2,a_3,a_4,a_5,a_6$ are the roots of the equation.
Then $a_1+a_2+a_3+a_4+a_5+a_6\geq 6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6} $
Now $a_1+a_2+a_3+a_4+a_5+a_6=12$
And also $6(a_1\times a_2\times a_3\times a_4\times a_5\times a_6)^{1/6}=12 $
Hence equality condition of AM-GM inequality holds, which implies that all roots are equal, and you know the sum of the roots is $12$, hence all roots are equal to $2$, now find $a,b,c,d$.
Let $x_i$ be our roots.
Now, by AM-GM $$2=\frac{\sum\limits_{i=1}^6x_i}{6}\geq\sqrt[6]{\prod_{i=1}^6x_i}=2,$$ which says that all $x_i=2$.
Thus, $$a=2^2\cdot\frac{6(6-1)}{2}=60,$$ $$b=-2^3\cdot\frac{6(6-1)(6-2)}{6}=-160,$$ $$c=2^4\cdot\frac{6(6-1)(6-2)(6-3)}{24}=240$$ and $$d=-2^5\cdot\frac{6(6-1)(6-2)(6-3)(6-4)}{120}=-192.$$