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Question Number 151789 by mathdanisur last updated on 23/Aug/21

Answered by ArielVyny last updated on 23/Aug/21

$$wehave0⩽\lambda ⩽1\to 0⩽\lambda +a+b⩽1+a+b$$$$andweadmitthata+b⩾\lambda +1;b+c⩾\lambda +1$$$$c+a⩾\lambda +1(notethatabc=1)$$$$0⩽\frac{1}{1+a+b}⩽\frac{1}{\lambda +a+b}⩽\frac{1}{\lambda +2}$$$$0⩽\frac{1}{1+b+c}⩽\frac{1}{\lambda +b+c}⩽\frac{1}{\lambda +2}$$$$0⩽\frac{1}{1+c+a}⩽\frac{1}{\lambda +c+a}⩽\frac{1}{\lambda +2}$$$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}⩽\frac{1}{\lambda +a+b}+\frac{1}{\lambda +b+c}+\frac{1}{\lambda +c+a}⩽\frac{1}{\lambda +2}+\frac{1}{\lambda +2}+\frac{1}{\lambda +2}$$$$then\frac{1}{\lambda +a+b}+\frac{1}{\lambda +b+c}+\frac{1}{\lambda +c+a}⩽\frac{3}{\lambda +2}$$

Commented by mathdanisur last updated on 23/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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