Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

5.53 Proof: The roots x1,x2x_{1}, x_{2} of the polynomial
x2+px−12p2,p∈R,p≠0x^{2}+p x-\frac{1}{2 p^{2}}, p \in R, p \neq 0

satisfy
x14+x24⩾2+2. x_{1}^{4}+x_{2}^{4} \geqslant 2+\sqrt{2} \text {. }

Solution

[Proof] According to Vieta's formulas,
x1+x2=−p,x1x2=−12p2x_{1}+x_{2}=-p, x_{1} x_{2}=-\frac{1}{2 p^{2}}

and the inequality between the arithmetic mean and the geometric mean of two numbers, we get
x14+x24=(x1+x2)4−2x1x2[2(x1+x2)2−x1x2]=p4+1p2(2p2+12p2)=p4+2+12p4⩾2+2p4⋅12p4=2+2,\begin{aligned} x_{1}^{4}+x_{2}^{4} & =\left(x_{1}+x_{2}\right)^{4}-2 x_{1} x_{2}\left[2\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}\right] \\ & =p^{4}+\frac{1}{p^{2}}\left(2 p^{2}+\frac{1}{2 p^{2}}\right) \\ & =p^{4}+2+\frac{1}{2 p^{4}} \\ & \geqslant 2+2 \sqrt{p^{4} \cdot \frac{1}{2 p^{4}}} \\ & =2+\sqrt{2}, \end{aligned}

which is what we needed to prove.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.