- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,587

Show $20a^2+20b^2+5c^2\ge 64$ if $y=x^4+ax^3+bx^2+cx+4$ has a real root.

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,587

Show $20a^2+20b^2+5c^2\ge 64$ if $y=x^4+ax^3+bx^2+cx+4$ has a real root.

- Aug 30, 2012

- 1,120

Made an oopsie!

-Dan

-Dan

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,587

$ax^3+bx^2+cx=-(x^4+4)\\(ax^3+bx^2+cx)^2=-(x^4+4)^2\\\left(2a\dfrac{x^3}{2}+2b\dfrac{x^2}{2}+cx \right)=-(x^4+4)^2 \le (4a^2+4b^2+c^2)\left(\dfrac{x^6}{4}+\dfrac{x^4}{4}+x^2\right)$

by the

This gives

$4a^2+4b^2+c^2\ge \dfrac{4(x^4+4)^2}{x^6+x^4+4x^2}$

Let $t=x^2$, now, we have to prove $\dfrac{4(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{64}{5}$, i.e. $\dfrac{(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{16}{5}$.

This is true since $\dfrac{(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{16}{5}$ implies $5(t^4+8t^2+16)\ge 16t^3+16t^2+64t$, or $(t-2)^2(5t^2+4t+20)\ge 0$.