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Question-152791




Question Number 152791 by mathdanisur last updated on 01/Sep/21
Answered by MJS_new last updated on 02/Sep/21
(xy−1)^2 −16(x−1)(y−1)≥0  x^2 y^2 −18xy+16x+16y−15≥0  x=3+p∧y=3+q∧p≥0∧q≥0  p^2 q^2 +6pq(p+q+3)+9(p^2 +q^2 )+16(p+q)≥0  which obviously is true
$$\left({xy}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{16}\left({x}−\mathrm{1}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{18}{xy}+\mathrm{16}{x}+\mathrm{16}{y}−\mathrm{15}\geqslant\mathrm{0} \\ $$$${x}=\mathrm{3}+{p}\wedge{y}=\mathrm{3}+{q}\wedge{p}\geqslant\mathrm{0}\wedge{q}\geqslant\mathrm{0} \\ $$$${p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{6}{pq}\left({p}+{q}+\mathrm{3}\right)+\mathrm{9}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{16}\left({p}+{q}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{obviously}\:\mathrm{is}\:\mathrm{true} \\ $$

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