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Algebra Difficulty 6.6 National olympiad Prove it

Inference 1 The necessary and sufficient condition for the symmetric cubic inequality satisfying P3(1,1,1)=0P_{3}(1,1,1)=0
P3(x,y,z)=λ0,1σ13+λ0,2σ1σ2+λ1,1σ30P_{3}(x, y, z)=\lambda_{0,1} \sigma_{1}^{3}+\lambda_{0,2} \sigma_{1} \sigma_{2}+\lambda_{1,1} \sigma_{3} \geq 0

is
P3(1,0,0)=λ0.10,P3(1,1,0)=2(4λ0,1+λ0,2)0P_{3}(1,0,0)=\lambda_{0.1} \geq 0, \quad P_{3}(1,1,0)=2\left(4 \lambda_{0,1}+\lambda_{0,2}\right) \geq 0 \text {. }

Solution

Proof: The necessity is obviously true. Now we prove the sufficiency.
Since P3(1,1,1)=0P_{3}(1,1,1)=0, according to the aforementioned method Schp, we can obtain
P3(x,y,z)=α0,1f0.1(3)+α0,2f0,2(3)=λ0.1f0.1(3)+(4λ0,1+λ0,2)f0.2(3),P_{3}(x, y, z)=\alpha_{0,1} f_{0.1}^{(3)}+\alpha_{0,2} f_{0,2}^{(3)}=\lambda_{0.1} f_{0.1}^{(3)}+\left(4 \lambda_{0,1}+\lambda_{0,2}\right) f_{0.2}^{(3)},

Therefore, if equation (3.1.2) holds, and by Theorem 2 we know f0,1(3)0,f0,2(3)0f_{0,1}^{(3)} \geq 0, f_{0,2}^{(3)} \geq 0, then equation (3.1.2) must hold. Q.E.D.

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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.