Question

Let x,y be positive real numbers, and m,n positive integers. The maximum value of the expression xᵐyⁿ(1−x²ᵐ)(1−y²ⁿ) is:

a) 0
b) 1
c) xy
d) 1/4​

Answer 1

The maximum value of the given expression is option d)) 1/4.

Step-by-step explanation:

To find the maximum value, we can use the AM-GM inequality.

The AM-GM inequality states that for any positive real numbers a and b, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM), i.e., AM ≥ GM.

Let's apply this inequality to the given expression.

1. By applying AM-GM inequality on xm and (1−x²ᵐ), we have:
   AM ≥ GM
   (xm + (1−x²ᵐ))/2 ≥ √(xm * (1−x²ᵐ))
   (1−x²ᵐ)/2 ≥ √(xm * (1−x²ᵐ))
2. Similarly, by applying AM-GM inequality on yn and (1−y²ⁿ), we have:
   (1−y²ⁿ)/2 ≥ √(yn * (1−y²ⁿ))
3. Multiplying the inequalities obtained from steps 1 and 2:
   ((1−x²ᵐ)/2) * ((1−y²ⁿ)/2) ≥ √(xm * (1−x²ᵐ) * yn * (1−y²ⁿ))
4. Substituting the given expression in step 3:
   ((1−x²ᵐ)/2) * ((1−y²ⁿ)/2) ≥ √(xm * (1−x²ᵐ) * yn * (1−y²ⁿ))
5. Taking the square root of both sides:
   √(((1−x²ᵐ)/2) * ((1−y²ⁿ)/2)) ≥ √(√(xm * (1−x²ᵐ) * yn * (1−y²ⁿ)))
   (√(1−x²ᵐ)/√2) * (√(1−y²ⁿ)/√2) ≥ xy
6. Rearranging the inequality:
   xᵐyⁿ * (1−x²ᵐ) * (1−y²ⁿ) ≤ 1/4 * (1/2 * (1/2))
   xᵐyⁿ * (1−x²ᵐ) * (1−y²ⁿ) ≤ 1/4

Hence, the maximum value of the expression xmyn(1−x²ᵐ)(1−y²ⁿ) is 1/4, which corresponds to option (d).