Tangent Line method helps here.

arqady wrote: Tangent Line method helps here. whats the minimum value arqady?

$2$ of course.

$(5a+9)(a-1)^2\geq0 \implies \frac {a+1}{a(a+2)}\geq -\frac {5}{9}(a-1)+\frac {2}{3},$ $\frac {a+1}{a(a+2)}+ \frac {b+1}{b(b+2)}+\frac {c+1}{c(c+2)} \geq-\frac {5}{9}(a+b+c-3)+2\geq2.$ Equality holds for $a=b=c=1$.

$\frac {a+1}{a(a+2)}+ \frac {b+1}{b(b+2)}+\frac {c+1}{c(c+2)}$ $=\frac {1}{2}\left( \frac {1}{a}+\frac {1}{b}+\frac {1}{c}+\frac {1}{a+2}+\frac {1}{b+2}+\frac {1}{c+2}\right)$ $\geq\frac {1}{2}\left( \frac {9}{a+b+c}+\frac {9}{a+b+c+6}\right)\geq2.$

Let $a_1,a_2,\cdots,a_n$ $(n\geq2)$ be positive real numbers such that $a_1+a_2+\cdots+a_n\leq n$.Prove that $$\frac {a_1+1}{a_1(a_1+n-1)}+ \frac {a_2+1}{a_2(a_2+n-1)}+ \cdots+\frac {a_n+1}{a_n(a_n+n-1)}\geq2.$$

sqing wrote: $\frac {a+1}{a(a+2)}+ \frac {b+1}{b(b+2)}+\frac {c+1}{c(c+2)}$ $=\frac {1}{2}\left( \frac {1}{a}+\frac {1}{b}+\frac {1}{c}+\frac {1}{a+2}+\frac {1}{b+2}+\frac {1}{c+2}\right)$ $\geq\frac {1}{2}\left( \frac {9}{a+b+c}+\frac {9}{a+b+c+6}\right)\geq2.$ nice! same as mine

sqing wrote: Let $a_1,a_2,\cdots,a_n$ $(n\geq2)$ be positive real numbers such that $a_1+a_2+\cdots+a_n\leq n$.Prove that $$\frac {a_1+1}{a_1(a_1+n-1)}+ \frac {a_2+1}{a_2(a_2+n-1)}+ \cdots+\frac {a_n+1}{a_n(a_n+n-1)}\geq2.$$ $\frac {a_1+1}{a_1(a_1+n-1)}+ \frac {a_2+1}{a_2(a_2+n-1)}+ \cdots+\frac {a_n+1}{a_n(a_n+n-1)}$ $=\frac {1}{n-1}\left( \frac {1}{a_1}+\frac {1}{a_2}+\cdots\frac {1}{a_n}\right)+\frac {n-2}{n-1}\left( \frac {1}{a_1+n-1}+\frac {1}{a_2+n-1}+\cdots\frac {1}{a_n+n-1}\right)$ $\geq\frac {1}{n-1}\cdot\frac {n^2}{a_1+a_2+\cdots+a_n}+\frac {n-2}{n-1}\cdot\frac {n^2}{a_1+a_2+\cdots+a_n+n(n-1)}$ $\geq\frac {1}{n-1}\cdot\frac {n^2}{ n}+\frac {n-2}{n-1}\cdot\frac {n^2}{n+n(n-1)}=2.$

$\frac{x(x+2)}{x+1}=x+1-\frac{1}{x+1}$ $\sum \frac{1}{a+1} \ge \frac{9}{3+ \sum a} \ge \frac{9}{6}=\frac{3}{2}$ $\sum \frac{a+1}{a(a+2)} \ge \frac{9}{\sum \frac{a(a+2)}{a+1}}=\frac{9}{3 + \sum a - \sum \frac{1}{a+1}} \ge \frac{9}{3+3-\frac{3}{2}}=\frac{9}{\frac{9}{2}}=2$ We will get the minimum for $a=b=c$ and when $a+b+c$ has it's maximum value and that $3$ so $a=b=c=1$.

gobathegreat wrote: Determine minimum value of the following expression: $\frac {a+1}{a(a+2)}+ \frac {b+1}{b(b+2)}+\frac {c+1}{c(c+2)}$ for positive real numbers such that $a+b+c \leq 3$ My solution : We divide the problem into two cases : Case 1 : If $\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\geq 1$ we write the given inequality : $$\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}+\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\geq 2$$ This is true since : $$\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\geq 1$$ and $$\dfrac{1}{a+2}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\geq \dfrac{9}{a+b+c+6}\geq 1$$ Case 2 : If $\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\leq 1$ we write the given inequality : $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq 2+\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}$$ This also is true since : $$\dfrac{1}{a(a+2)}+\dfrac{1}{b(b+2)}+\dfrac{1}{c(c+2)}\leq 1$$ and $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{9}{a+b+c}\geq 3$$

arqady wrote: Tangent Line method helps here. Yeh u are right .

A=(a+1)/(a(a+2))+(b+1)/(b(b+2))+(c+1)/(c(c+2)) (a+2)/(a(a+2))=1/2 ((2a+2)/a(a+2) )=1/2(1/a+1/(a+2)) A=1/2 ∑▒1/a+1/(a+2) ∑▒a ∑▒1/a≥9 → ∑▒1/a≥3 ∑▒〖a+2〗 ∑▒1/(a+2)≥9 → ∑▒1/(a+2)≥1 A=1/2 ∑▒1/a+∑▒1/(a+2)≥(3+1)/2=2 the equality:a=b=c=1