Question Number 143591 by mathdanisur last updated on 16/Jun/21
$${a};{b};{c}>\mathrm{0}\:\:{and}\:\:{a}+{b}+{c}={k} \\ $$$$\boldsymbol{{min}}\left(\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{b}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}+{c}^{\mathrm{2}} }\right)=? \\ $$
Answered by Olaf_Thorendsen last updated on 16/Jun/21
$$\mathrm{Let}\:{f}\left({a},{b},{c}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{\mathrm{2}} } \\ $$$${min}\left({f}\left({a},{b},{c}\right)\right)\:=\:? \\ $$$$\mathrm{By}\:\mathrm{symmetry}\:{a}\:=\:{b}\:=\:{c}\:=\:\frac{{k}}{\mathrm{3}} \\ $$$${min}\left({f}\left({a},{b},{c}\right)\right)\:=\:\frac{\mathrm{3}}{\mathrm{1}+\left(\frac{{k}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$${min}\left({f}\left({a},{b},{c}\right)\right)\:=\:\frac{\mathrm{27}}{\mathrm{9}+{k}^{\mathrm{2}} } \\ $$
Commented by mathdanisur last updated on 16/Jun/21
$${thanks}\:{Sir},\:{there}\:{are}\:{also}\:{two} \\ $$$${other}\:{roots}.. \\ $$