Question Number 198147 by mr W last updated on 11/Oct/23
$${if}\:{a},{x},{y},{b}\:{is}\:{an}\:{AP}\:{and}\:{a},{p},{q},{b}\:{is}\:{a}\:{GP}. \\ $$$${prove}\:{that}\:{xy}\geqslant{pq}. \\ $$$$\left({with}\:{a},\:{b}\:>\mathrm{0}\right) \\ $$
Answered by AST last updated on 11/Oct/23
$${p}^{\mathrm{2}} ={aq};{q}^{\mathrm{2}} ={pb}\Rightarrow{p}^{\mathrm{2}} {q}^{\mathrm{2}} ={abpq}\Rightarrow{pq}={ab} \\ $$$$\mathrm{2}{x}={a}+{y};\mathrm{2}{y}={x}+{b}\Rightarrow\mathrm{4}{xy}=\left({a}+{y}\right)\left({x}+{b}\right) \\ $$$$\mathrm{3}{xy}={ax}+{ab}+{yb}\overset{?} {\geqslant}\mathrm{3}{pq}=\mathrm{3}{ab} \\ $$$$\Leftrightarrow{ax}+{yb}\geqslant\mathrm{2}{ab} \\ $$$$\Leftrightarrow{a}\left({a}+{d}\right)+\left({a}+\mathrm{2}{d}\right)\left({a}+\mathrm{3}{d}\right)\geqslant\mathrm{2}\left({a}\right)\left({a}+\mathrm{3}{d}\right) \\ $$$${a}^{\mathrm{2}} +{ad}+{a}^{\mathrm{2}} +\mathrm{3}{ad}+\mathrm{2}{ad}+\mathrm{6}{d}^{\mathrm{2}} \geqslant\mathrm{2}{a}^{\mathrm{2}} +\mathrm{6}{ad} \\ $$$$\Leftrightarrow{d}^{\mathrm{2}} \geqslant\mathrm{0}\left({true}\right)..\:{Hence}\:\mathrm{3}{xy}\geqslant\mathrm{3}{ab}\Rightarrow{xy}\geqslant{ab}={pq} \\ $$$${Equality}\:{holds}\:{when}\:{d}=\mathrm{0}\:\Rightarrow\:{a}={b}={x}={y}={p}={q} \\ $$
Commented by mr W last updated on 11/Oct/23
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