Key Concepts and Formulas

• Arithmetic Mean-Geometric Mean (AM-GM) Inequality: For non-negative real numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$. Equality holds when $a=b$. A useful form is $a+b \ge 2\sqrt{ab}$.
• Special Case of AM-GM for Reciprocals: For a positive real number $A$, $A + \frac{1}{A} \ge 2$. Equality holds when $A=1$.

Step-by-Step Solution

Step 1: Analyze the Expression and Identify the Strategy We are asked to find the maximum value of the expression $E = {{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}}$, where $x, y > 0$ and $m, n$ are positive integers. To maximize this fraction, we aim to minimize its denominator. The presence of terms like $(1 + x^{2m})$ and $(1 + y^{2n})$ in the denominator suggests using the AM-GM inequality.

Step 2: Rewrite the Expression to Apply AM-GM To effectively use AM-GM, we can manipulate the expression by dividing the numerator and the denominator by $x^m y^n$. $E = {{{x^m}{y^n}} \over {\left( {1 + {x^{2m}}} \right)\left( {1 + {y^{2n}}} \right)}} = {1 \over {{{{1 + {x^{2m}}} \over {{x^m}}}} {{{1 + {y^{2n}}} \over {{y^n}}}}}}$ Now, let's simplify the terms in the denominator: ${{{1 + {x^{2m}}} \over {{x^m}}}} = {1 \over {{x^m}}} + {{{x^{2m}}} \over {{x^m}}} = {x^{-m}} + {x^m}$ And similarly for the $y$ term: ${{{1 + {y^{2n}}} \over {{y^n}}}} = {1 \over {{y^n}}} + {{{y^{2n}}} \over {{y^n}}} = {y^{-n}} + {y^n}$ Substituting these back into the expression for $E$, we get: $E = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}}$ Reasoning: This rearrangement is key because it transforms the terms in the denominator into the form $A + \frac{1}{A}$, which is directly amenable to the AM-GM inequality.

Step 3: Apply AM-GM to the Denominator Terms Consider the term $\left( {{x^m} + {1 \over {{x^m}}}} \right)$. Since $x > 0$ and $m$ is a positive integer, $x^m > 0$. Let $A = x^m$. Applying the AM-GM inequality $A + \frac{1}{A} \ge 2$: ${x^m} + {1 \over {{x^m}}} \ge 2\sqrt{{x^m} \cdot {1 \over {{x^m}}}} = 2\sqrt{1} = 2$ So, ${x^m} + {1 \over {{x^m}}} \ge 2$. (Inequality 1) Reasoning: We use the special case of AM-GM for reciprocals to find the minimum possible value of this part of the denominator.

Similarly, consider the term $\left( {{y^n} + {1 \over {{y^n}}}} \right)$. Since $y > 0$ and $n$ is a positive integer, $y^n > 0$. Let $B = y^n$. Applying AM-GM: ${y^n} + {1 \over {{y^n}}} \ge 2\sqrt{{y^n} \cdot {1 \over {{y^n}}}} = 2\sqrt{1} = 2$ So, ${y^n} + {1 \over {{y^n}}} \ge 2$. (Inequality 2) Reasoning: The same AM-GM application is used for the $y$ dependent term.

Step 4: Combine Inequalities to Bound the Denominator From Inequality 1 and Inequality 2, we have: $\left( {{x^m} + {1 \over {{x^m}}}} \right) \ge 2$ $\left( {{y^n} + {1 \over {{y^n}}}} \right) \ge 2$ Since both factors in the denominator are positive, we can multiply these inequalities: $\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right) \ge 2 \cdot 2 = 4$ Reasoning: The product of two positive lower bounds gives a lower bound for the product. This establishes the minimum value of the entire denominator.

Step 5: Determine the Maximum Value of the Expression Now, we use the bound for the denominator in our expression for $E$: $E = {1 \over {\left( {{x^m} + {1 \over {{x^m}}}} \right)\left( {{y^n} + {1 \over {{y^n}}}} \right)}}$ Since the denominator is greater than or equal to 4, the reciprocal (which is $E$) will be less than or equal to $\frac{1}{4}$: $E \le {1 \over 4}$ Reasoning: As the denominator's value increases, the fraction's value decreases, and vice versa. To find the maximum value of $E$, we use the minimum value of the denominator.

Step 6: Check for Equality The maximum value of $\frac{1}{4}$ is achieved when equality holds in both AM-GM applications. Equality in ${x^m} + {1 \over {{x^m}}} \ge 2$ holds when $x^m = {1 \over {{x^m}}}$, which implies $(x^m)^2 = 1$. Since $x > 0$, we must have $x^m = 1$, which means $x=1$. Equality in ${y^n} + {1 \over {{y^n}}} \ge 2$ holds when $y^n = {1 \over {{y^n}}}$, which implies $(y^n)^2 = 1$. Since $y > 0$, we must have $y^n = 1$, which means $y=1$. Let's substitute $x=1$ and $y=1$ into the original expression: $E = {{{1^m}{1^n}} \over {\left( {1 + {1^{2m}}} \right)\left( {1 + {1^{2n}}} \right)}} = {{1} \over {\left( {1 + 1} \right)\left( {1 + 1} \right)}} = {1 \over {2 \cdot 2}} = {1 \over 4}$ Since the value $\frac{1}{4}$ is attainable, it is indeed the maximum value.

Common Mistakes & Tips

• Incorrect Application of AM-GM: Ensure you are applying AM-GM to positive terms. Incorrectly applying it to negative terms will lead to wrong results.
• Forgetting the Equality Condition: Always check if the maximum/minimum value derived from AM-GM is actually achievable by finding the values of the variables that satisfy the equality condition.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with exponents and fractions. Rewriting the expression as shown in Step 2 is a common and effective technique for problems of this type.

Summary

The problem requires finding the maximum value of a given expression involving positive real numbers and positive integers. By strategically rewriting the expression, we were able to apply the AM-GM inequality to the terms in the denominator. The AM-GM inequality, specifically the form $A + \frac{1}{A} \ge 2$, was used to find the minimum values of the individual factors in the denominator. Multiplying these minimum values gave the minimum value of the entire denominator. The reciprocal of this minimum denominator value yielded the maximum value of the original expression. The condition for equality in AM-GM confirmed that this maximum value is indeed attainable.

The final answer is $\boxed{{1 \over 4}}$.