Question Number 151890 by mathdanisur last updated on 23/Aug/21
Answered by mnjuly1970 last updated on 24/Aug/21
$$\:\:\:{x}^{\:\mathrm{2}} +{y}^{\:\mathrm{2}} +{z}^{\:\mathrm{2}} \:+\left({x}^{\:\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\right)+\left({y}^{\:\mathrm{2}} +\frac{\mathrm{1}}{{y}^{\:\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:+\:\left(\:{z}^{\:\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\:\mathrm{2}} }\:\right)\:+\mathrm{2}\:\left(\:\frac{{x}}{{y}}\:+\frac{{y}}{{z}}\:+\frac{{z}}{{x}}\:\right) \\ $$$$\:\:\geqslant\:\left({xy}\:+{xz}\:+{yz}\:\right)+\mathrm{6}\:+\mathrm{6}\:+\mathrm{6}=\mathrm{19} \\ $$
Commented by mathdanisur last updated on 24/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser},\:\mathrm{but}\:=\:\mathrm{16}\:\mathrm{or}\:\mathrm{19}.? \\ $$