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

[anemone]

Let $a$ amd $b$ be positive reals such that $a+b=ab$.

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

 

[Greg]

Let $a$ amd $b$ be positive reals such that $a+b=ab$.

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

Spoiler

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
$$\min\left(\dfrac{a}{b^2+4}+\dfrac{b}{a^2+4}\right)=2\sqrt{\dfrac{2\cdot2}{(4+4)(4+4)}}=\dfrac12$$
$$\text{Q. E. D.}$$

 

[MarkFL]

My solution:

Spoiler

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:

$$f_{\min}=\frac{1}{2}$$

 

[anemone]

Thanks greg1313 and MarkFL for participating and the neat and well-written solution! Bravo!(Cool)

My solution:

Spoiler

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