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Question Number 144823 by loveineq last updated on 29/Jun/21
Let a,b > 0 and a+b+1 = 3ab. Prove that ((a+1)/(b+1))+((b+1)/(a+1)) ≤ a+b
$$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+\mathrm{1}\:=\:\mathrm{3}{ab}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\:\leqslant\:{a}+{b} \\ $$
Answered by ArielVyny last updated on 30/Jun/21
we know that a+b+1=3ab we suppose ab≥1 (a+1)=3ab−b (a+1)^2 =b^2 (3a−1)^2 =b^2 (9a^2 +1−9a) a+1≤9ab (1) →b+1≤9ab (2) (((1))/((2)))+(((2))/((1)))→((a+1)/(b+1))+((b+1)/(a+1))≤2 then 3≤3ab→2≤3ab−1 we have ((a+1)/(b+1))+((b+1)/(a+1))≤2≤3ab−1=a+b ((a+1)/(b+1))+((b+1)/(a+1))≤2≤a+b finally ((a+1)/(b+1))+((b+1)/(a+1))≤a+b (ab≥1)
$${we}\:{know}\:{that}\:{a}+{b}+\mathrm{1}=\mathrm{3}{ab} \\ $$$${we}\:{suppose}\:{ab}\geqslant\mathrm{1} \\ $$$$\left({a}+\mathrm{1}\right)=\mathrm{3}{ab}−{b} \\ $$$$\left({a}+\mathrm{1}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{3}{a}−\mathrm{1}\right)^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{9}{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{9}{a}\right) \\ $$$${a}+\mathrm{1}\leqslant\mathrm{9}{ab}\:\left(\mathrm{1}\right)\:\:\:\rightarrow{b}+\mathrm{1}\leqslant\mathrm{9}{ab}\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)}{\left(\mathrm{2}\right)}+\frac{\left(\mathrm{2}\right)}{\left(\mathrm{1}\right)}\rightarrow\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\leqslant\mathrm{2} \\ $$$${then}\:\:\mathrm{3}\leqslant\mathrm{3}{ab}\rightarrow\mathrm{2}\leqslant\mathrm{3}{ab}−\mathrm{1} \\ $$$${we}\:{have}\:\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\leqslant\mathrm{2}\leqslant\mathrm{3}{ab}−\mathrm{1}={a}+{b} \\ $$$$\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\leqslant\mathrm{2}\leqslant{a}+{b} \\ $$$${finally}\:\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}+\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\leqslant{a}+{b}\:\:\:\:\left({ab}\geqslant\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$

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