$a+b=ab$. $\implies{ab\geq{4}}$ from cauchy's S $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}= \frac{a^2}{a b^{2}+4 a}+\frac{b^2}{b a^{2}+4b}\geq{\frac{(a+b)^2}{ab(a+b)+4(a+b)}}=\frac{1}{1+\frac{4}{ab}}\geq{\frac{1}{2}}$

Replace $a,b$ with $\frac{1}{x},\frac{1}{y}$ so that $a+b=ab\implies x+y=\frac{1}{a}+\frac{1}{b}=1$. Then $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}=\frac{y^2}{4xy^2+x}+\frac{x^2}{4yx^2+y}=T$. By CS $(x+y+4xy^2+4yx^2)T\ge (x+y)^2=1$. Now $x+y+4xy^2+4yx^2=4xy+1$ so $T\ge\frac{1}{4xy+1}$ so it suffices to prove $2\ge 4xy+1$ i.e. $xy \le \frac{1}{4}=\frac{(x+y)^2}{4}$, true by AM-GM.

$(a+b)^2 \geq 4ab = 4(a+b)$ we have $a+b \geq 4$ $\frac{a}{b^{2}+4}= \frac{1}{4}a(1-\frac{b^2}{b^{2}+4}) \geq \frac{1}{4}a(1-\frac{b^2}{4b})= \frac{1}{4}a(1-\frac{b}{4})= \frac{1}{16}(4a-ab)$ The same $\frac{b}{a^{2}+4}=\geq \frac{1}{16}(4b-ab)$ we have $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}\geq \frac{1}{16}(4a +4b-2ab)=\frac{1}{8}(a+b) \geq\frac{1}{2}$. Teaching Mathematics for High School for Gifted Students, Hanoi National University of Education Vietnam http://www.artofproblemsolving.com/Foru ... =egosearch [Moderator edit: Moved post from this topic]

actually it is very easy problem however here is my solution!) $a+b=ab=>ab\ge4$ $\frac {a} {b^2+4}+\frac {b} {a^2+4}\ge \frac {(a+b)^2} {(a+b)(ab+4)}=\frac{ab} {ab+4}\ge \frac {1} {2}$ which is obvious!)

momo1729 wrote: Let $a$ and $b$ be two positive real numbers such that $a+b=ab$. Prove that $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}\geq \frac{1}{2}$.

From MA-MG we have that $ab \ge 4$ Then $ \frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}\ge \frac{a}{b^{2}+ab}+\frac{b}{a^{2}+ab} = \frac{a}{b(a+b)}+\frac{b}{a(a+b)} = \frac{1}{a^2} + \frac{1}{b^2}$ So it suffices to prove $\frac{1}{a^2} + \frac{1}{b^2} \ge \frac {1}{2}$ $\leftrightarrow \frac{a^2+b^2}{a^2b^2} \ge \frac{1}{2}$ $\leftrightarrow 2(a^2+b^2) \ge (a+b)^2$ which is obviously true.

$(a-1)(b-1)=1$, so $a-1=\frac{x}{y},b-1=\frac{y}{x},a=\frac{x+y}{y},b=\frac{x+y}{y}$,$x,y>0$, so the inequality is equivalent (after clearing the denomitors )to $2(x^6+y^6)+2x^2y^2(x^2+y^2)+xy^5+x^5y\ge 10x^3y^3$, obvious by AM-GM.

Generalization. Let $a$ and $b$ be two positive real numbers such that $a+b=ab$.Prove that\[\frac{a}{b^{2}+\lambda }+\frac{b}{a^{2}+\lambda }\geq \frac{4}{4+\lambda }(\lambda \ge 0)\].

denote $ \frac{1}{a}=x,\frac{1}{b}=y $ , then $ x+y=1\Rightarrow 1\geq 2\sqrt{xy}\Rightarrow 4xy\leq 1, (*) $, we $ \frac{a}{b^{2}+\lambda }+\frac{b}{a^{2}+\lambda}=\frac{y^{2}}{x+xy^{2}\lambda}+\frac{x^{2}}{y+yx^{2}\lambda}\geq \frac{(x+y)^{2}}{x+y+\lambda xy(x+y)}=\frac{1}{1+\lambda xy}\geq \frac{4}{4+\lambda}\Leftrightarrow 4xy\leq 1, (*) $ it's true

momo1729 wrote: Let $a$ and $b$ be two positive real numbers such that $a+b=ab$. Prove that $\frac{a}{b^{2}+4}+\frac{b}{a^{2}+4}\geq \frac{1}{2}$. sqing wrote: Generalization. Let $a$ and $b$ be two positive real numbers such that $a+b=ab$.Prove that\[\frac{a}{b^{2}+\lambda }+\frac{b}{a^{2}+\lambda }\geq \frac{4}{4+\lambda }(\lambda \ge 0)\]. Generalization Let $a$ and $b$ be two positive real numbers and $\lambda \ge 0$. Prove that $\frac{a}{{{b}^{2}}+\lambda }+\frac{b}{{{a}^{2}}+\lambda }\ge \frac{4ab{{\left( a+b \right)}^{2}}}{4{{a}^{2}}{{b}^{2}}\left( a+b \right)+\lambda {{\left( a+b \right)}^{3}}}+\frac{4\lambda \left( a+b \right){{\left( a-b \right)}^{2}}}{ab{{\left( 4ab+\lambda \left( a+b \right) \right)}^{2}}}=$ $\frac{4{{\left( \frac{a+b}{ab} \right)}^{2}}}{4\left( \frac{a+b}{ab} \right)+\lambda {{\left( \frac{a+b}{ab} \right)}^{3}}}+\frac{4\lambda \frac{a+b}{ab}{{\left( \frac{1}{a}-\frac{1}{b} \right)}^{2}}}{{{\left( 4+\lambda \frac{a+b}{ab} \right)}^{2}}}$ If $a+b=ab\Rightarrow \frac{a}{{{b}^{2}}+\lambda }+\frac{b}{{{a}^{2}}+\lambda }\ge \frac{4}{4+\lambda }+\frac{4\lambda {{\left( \frac{1}{a}-\frac{1}{b} \right)}^{2}}}{{{\left( 4+\lambda \right)}^{2}}}$ If $a+b=\mu ab\Rightarrow \frac{a}{{{b}^{2}}+\lambda }+\frac{b}{{{a}^{2}}+\lambda }\ge \frac{4{{\mu }^{2}}}{4\mu +\lambda {{\mu }^{3}}}+\frac{4\lambda \mu {{\left( \frac{1}{a}-\frac{1}{b} \right)}^{2}}}{{{\left( 4+\lambda \mu \right)}^{2}}}$

$a+b=ab=>ab\ge4$ $\frac {a} {b^2+4}+\frac {b} {a^2+4}\ge \frac {(a+b)^2} {a(b^2+4)+b(a^2+4)}=\frac{ab} {ab+4}\ge 1-\frac{4} {ab+4}\ge 1-\frac{1} {2}=\frac {1} {2}.$