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Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$

I tried doing it with CS-Engel to get $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c} $$ I thought that maybe proof that $$ \frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)} $$ or $$ 3+3 a b c \geq a+b+c+a b+b c+a c $$, but I don’t know how

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Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$

I tried doing it with CS-Engel to get $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c} $$ I thought that maybe proof that $$ \frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)} $$ or $$ 3+3 a b c \geq a+b+c+a b+b c+a c $$, but I don’t know how

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Proof that 1/(a(1+b))+1/(b(1+c))+1/(c(1+a))>=3/(1+abc)

I tried doing it with CS-Engel to get $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c} $$ I thought that maybe proof that $$ \frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)} $$ or $$ 3+3 a b c \geq a+b+c+a b+b c+a c $$, but I don’t know how