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Question Number 150889 by mathdanisur last updated on 16/Aug/21

Answered by dumitrel last updated on 16/Aug/21

⇔a^2 +a(b+c+d)+(b+c+d)^2 ≥0⇔  ((a/2)+b+c+d)^2 +((3a^2 )/4)≥0  ,,=′′ for a=0; b+c+d=0

$$\Leftrightarrow{a}^{\mathrm{2}} +{a}\left({b}+{c}+{d}\right)+\left({b}+{c}+{d}\right)^{\mathrm{2}} \geqslant\mathrm{0}\Leftrightarrow \\ $$$$\left(\frac{{a}}{\mathrm{2}}+{b}+{c}+{d}\right)^{\mathrm{2}} +\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}\geqslant\mathrm{0} \\ $$$$,,=''\:{for}\:{a}=\mathrm{0};\:{b}+{c}+{d}=\mathrm{0} \\ $$

Commented by mathdanisur last updated on 16/Aug/21

Thank you Ser nice

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{nice} \\ $$

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