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

Answered by ArielVyny last updated on 23/Aug/21

we have 0≤λ≤1→0≤λ+a+b≤1+a+b and we admit that a+b≥λ+1;b+c≥λ+1 c+a≥λ+1 (note that abc=1) 0≤(1/(1+a+b))≤(1/(λ+a+b)) ≤(1/(λ+2)) 0≤(1/(1+b+c))≤(1/(λ+b+c))≤(1/(λ+2)) 0≤(1/(1+c+a))≤(1/(λ+c+a))≤(1/(λ+2)) (1/(1+a+b))+(1/(1+b+c))+(1/(1+c+a))≤(1/(λ+a+b))+(1/(λ+b+c))+(1/(λ+c+a))≤(1/(λ+2))+(1/(λ+2))+(1/(λ+2)) then (1/(λ+a+b))+(1/(λ+b+c))+(1/(λ+c+a))≤(3/(λ+2))

$${we}\:{have}\:\mathrm{0}\leqslant\lambda\leqslant\mathrm{1}\rightarrow\mathrm{0}\leqslant\lambda+{a}+{b}\leqslant\mathrm{1}+{a}+{b} \\ $$$${and}\:{we}\:{admit}\:{that}\:{a}+{b}\geqslant\lambda+\mathrm{1};{b}+{c}\geqslant\lambda+\mathrm{1} \\ $$$${c}+{a}\geqslant\lambda+\mathrm{1}\:\:\left({note}\:{that}\:{abc}=\mathrm{1}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\leqslant\frac{\mathrm{1}}{\lambda+{a}+{b}}\:\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{c}+{a}}\leqslant\frac{\mathrm{1}}{\lambda+{c}+{a}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}+\frac{\mathrm{1}}{\mathrm{1}+{c}+{a}}\leqslant\frac{\mathrm{1}}{\lambda+{a}+{b}}+\frac{\mathrm{1}}{\lambda+{b}+{c}}+\frac{\mathrm{1}}{\lambda+{c}+{a}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}}+\frac{\mathrm{1}}{\lambda+\mathrm{2}}+\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$${then}\:\frac{\mathrm{1}}{\lambda+{a}+{b}}+\frac{\mathrm{1}}{\lambda+{b}+{c}}+\frac{\mathrm{1}}{\lambda+{c}+{a}}\leqslant\frac{\mathrm{3}}{\lambda+\mathrm{2}} \\ $$

Commented by mathdanisur last updated on 23/Aug/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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