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Question Number 164856 by Zaynal last updated on 22/Jan/22

$$\mathbf{Let}\mathit{a},\mathit{b},\mathit{c}>0;$$ $$42\mathit{a}\mathit{b}\mathit{c}+4\mathit{b}\mathit{c}\mathit{a}+1\mathit{c}\mathit{a}\mathit{b}⩽24$$ $$\mathrm{T}\mathbf{hen}\frac{1}{\mathbf{ab}}+\frac{2}{\mathbf{bc}}+\frac{3}{\mathbf{ca}}=??$$ $${}^{\{Z.\mathrm{A}\}}$$

Commented byMJS_new last updated on 24/Jan/22

$$abc=bca=cab$$ $$\mathrm{anyway}\mathrm{you}\mathrm{cannot}\mathrm{solve}\mathrm{this}\mathrm{equation}\mathrm{with}$$ $$\mathrm{only}one\mathrm{inequation}\mathrm{given}$$

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