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Question Number 152314 by mathdanisur last updated on 27/Aug/21

Answered by ghimisi last updated on 27/Aug/21

(a^2 /(a^2 +ab+b^2 ))+(c^2 /(c(a+b+c)))≥(((a+c)^2 )/(a^2 +b^2 +c^2 +ab+bc+ac))  (b^2 /(b^2 +cb+c^2 ))+(a^2 /(a(a+b+c)))≥(((a+b)^2 )/(a^2 +b^2 +c^2 +ab+bc+ac))  (c^2 /(c^2 +ac+a^2 ))+(b^2 /(b(a+b+c)))≥(((b+c)^2 )/(a^2 +b^2 +c^2 +ab+bc+ac))⇒  Σ(a^2 /(a^2 +ab+b^2 ))+1≥2⇒Σ(a^2 /(a^2 +ab+b^2 ))≥1

$$\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{c}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}} \\ $$$$\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{cb}+{c}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{{a}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}} \\ $$$$\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{b}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}}\Rightarrow \\ $$$$\Sigma\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }+\mathrm{1}\geqslant\mathrm{2}\Rightarrow\Sigma\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }\geqslant\mathrm{1} \\ $$

Commented by mathdanisur last updated on 27/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$

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