Prove that $x_1^4+x_2^4\geq 2$

Question

If $x_1$ and $x_2$ are the solutions of the equation $x^2+px- \frac{a}{p^2}=0$, a=$\frac{1}{\sqrt{2}}+2$ and $p \in R-\{0\}$ Then prove that $x_1^4+x_2^4\geq 2$.

Can i go with Viet’s Formulas $x_1+x_2=-b/a$, then apply the power of 4 on both sides but I’m stuck here. Someone help please.

Answer 1

Note that \begin{align} x_1^4 + x_2^4 &= (x_1+x_2)^4 - 4x_1^3x_2-6x_1^2x_2^2-4x_1x_2^2 = \\ &= (x_1+x_2)^4 - x_1x_2 (4x_1^2+6x_1x_2+4x_2^2) = \\ &= (x_1+x_2)^4 - x_1x_2 \big(4(x_1+x_2)^2-2x_1x_2\big) \end{align} Then use the Vieta's formulae for $x_1+x_2$ and $x_1x_2$. See if you can show that what you get is greater of equal to $2$.