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Question Number 144676 by loveineq last updated on 27/Jun/21

$$\mathrm{Let}a,b,c>0\mathrm{and}(a+b)(b+c)=4.\mathrm{Prove}\mathrm{that}$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{b}{ca}⩾\frac{27}{8}$$ $$\left(\mathrm{Found}\mathrm{by}\mathrm{WolframAlpha}\right)$$

Answered by ArielVyny last updated on 27/Jun/21

$$$$ $$supposeabc⩽1$$ $$27abc⩽32$$ $$\frac{27}{8}abc⩽4$$ $$\frac{27}{8}⩽\frac{4}{abc}$$ $$or4=(a+b)(b+c)$$ $$\frac{27}{8}⩽\frac{cb+ac+ab+b^2}{abc}$$ $$\frac{27}{8}⩽\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{b}{ac}$$ $$then\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{b}{ac}⩾\frac{27}{8}$$ $$$$

Commented byloveineq last updated on 28/Jun/21

$$niceandthanks.$$

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