MHB Proving $\frac{a}{b^2+4}+\frac{b}{a^2+4}\ge \frac{1}{2}$ with $a+b=ab$

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The discussion revolves around proving the inequality $\frac{a}{b^2+4}+\frac{b}{a^2+4}\ge \frac{1}{2}$ under the condition that $a+b=ab$ for positive reals $a$ and $b$. Participants share their approaches and solutions, highlighting the importance of manipulating the given condition to derive the inequality. The conversation emphasizes the use of algebraic techniques and inequalities to establish the proof. The contributions from members enhance the understanding of the problem and its solution. The thread concludes with a sense of accomplishment in solving the inequality.
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Let $a$ amd $b$ be positive reals such that $a+b=ab$.

Prove that $$\frac{a}{b^2+4}+\frac{b}{a^2+4}\ge \frac{1}{2}$$.
 
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anemone said:
Let $a$ amd $b$ be positive reals such that $a+b=ab$.

Prove that $$\frac{a}{b^2+4}+\frac{b}{a^2+4}\ge \frac{1}{2}$$.

From the AM-GM inequality we have
$$\dfrac{a}{b^2+4}+\dfrac{b}{a^2+4}\ge2\sqrt{\dfrac{ab}{(a^2+4)(b^2+4)}}\quad(1)$$

We have equality when
$$\dfrac{a}{b^2+4}=\dfrac{b}{a^2+4}\implies a=b$$

With that equality and from $a+b=ab$, we derive $2a=a^2\implies a=b=2$.

Substituting this into $(1)$ we have
$$\min\left(\dfrac{a}{b^2+4}+\dfrac{b}{a^2+4}\right)=2\sqrt{\dfrac{2\cdot2}{(4+4)(4+4)}}=\dfrac12$$
$$\text{Q. E. D.}$$
 
My solution:

Let:

$$f(a,b)\equiv\frac{a}{b^2+4}+\frac{b}{a^2+4}$$

Using cyclic symmetry, we know the extremum occurs for:

$$a=b=2$$

Thus, the critical value is:

$$f(2,2)=\frac{2^2}{2^2+4}=\frac{1}{2}$$

If we pick another point on the constraint, such as:

$$(a,b)=\left(\frac{3}{2},3\right)$$

We find:

$$f\left(\frac{3}{2},3\right)=\frac{\dfrac{3}{2}}{3^2+4}+\frac{3}{\left(\dfrac{3}{2}\right)^2+4}=\frac{387}{650}>\frac{1}{2}$$

Hence, we may now assert:

$$f_{\min}=\frac{1}{2}$$
 
Thanks greg1313 and MarkFL for participating and the neat and well-written solution! Bravo!(Cool)

My solution:
$$\begin{align*}\frac{a}{b^2+4}+\frac{b}{a^2+4}&=\frac{a^2}{ab^2+4a}+\frac{b^2}{a^2b+4b}\\&\ge \frac{(a+b)^2}{ab^2+a^2b+4a+4b} \\&\ge \frac{(a+b)^2}{ab(a+b)+4(a+b)}\\& =\frac{a+b}{ab+4}\\&=\frac{ab}{ab+4}\,\,\,\text{since}\,\,\, a+b=ab\\&=\frac{1}{1+\frac{4}{ab}}\,\,\,\text{but}\,\,\, ab\ge 4\,\,\,\text{from}\,\,\, a+b=ab\ge 2\sqrt{ab}\\&\ge \frac{1}{1+\frac{4}{4}}=\frac{1}{2}\end{align*}$$
 
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