Question Number 152314 by mathdanisur last updated on 27/Aug/21
Answered by ghimisi last updated on 27/Aug/21
$$\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }+\frac{{c}^{\mathrm{2}} }{{c}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({a}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}} \\ $$$$\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} +{cb}+{c}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} }{{a}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({a}+{b}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}} \\ $$$$\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} }{{b}\left({a}+{b}+{c}\right)}\geqslant\frac{\left({b}+{c}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}+{bc}+{ac}}\Rightarrow \\ $$$$\Sigma\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }+\mathrm{1}\geqslant\mathrm{2}\Rightarrow\Sigma\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }\geqslant\mathrm{1} \\ $$
Commented by mathdanisur last updated on 27/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$