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Question Number 151350 by mathdanisur last updated on 20/Aug/21

Commented by dumitrel last updated on 20/Aug/21

$$0⩽\lambda ⩽8sau0⩽\lambda ⩽5?????$$$$p^2⩾3q\Rightarrow q⩽3$$$${(a+b+c)}^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)-3abc\Rightarrow$$$$a^3+b^3+c^3=27-9q+3r$$$$schur\Rightarrow p^3+9r⩾4pq\Rightarrow 4q⩽9+3r(\ast )$$$$\iff 27-9q+3r+\lambda q⩾3\lambda +3\iff \lambda (3-q)⩽24-9q+3r$$$$q=3true$$$$q<3\iff \lambda ⩽\frac{24-9q+3r}{3-q}(\bullet )$$$$\frac{24-9q+3r}{3-q}⩾5\iff 4q⩽9+3r\iff (\ast )$$$$0⩽\lambda ⩽5;\frac{24-9q+3r}{3-q}⩾5\Rightarrow (\bullet )true$$

Commented by mathdanisur last updated on 20/Aug/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er},\mathrm{Cool}$$

Commented by mathdanisur last updated on 20/Aug/21

$$\mathrm{true}\mathrm{Dear}\mathbf{S}\mathrm{er},0⩽\lambda ⩽5$$

Commented by dumitrel last updated on 20/Aug/21

$$5<\lambda ⩽8thisistrue?$$

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