Numerical answer
x and y are positive real numbers with x + y = 10. What is the minimum possible value of x² + 4y²?
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Answer: 80
Substitute y = 10 − x:
x² + 4(10 − x)² = 5x² − 80x + 400.
This quadratic is minimised at x = 80/(2 × 5) = 8, giving y = 2 and value 64 + 16 = 80.
Alternative (Cauchy–Schwarz): (x² + 4y²)(1 + 1/4) ≥ (x + y)² = 100 ⇒ x² + 4y² ≥ 100 × 4/5 = 80, with equality when x/1 = 2y/(1/2), i.e. x = 4y.
Tip: For a sum constraint, the minimum of ax² + by² occurs when ax = by.
x² + 4(10 − x)² = 5x² − 80x + 400.
This quadratic is minimised at x = 80/(2 × 5) = 8, giving y = 2 and value 64 + 16 = 80.
Alternative (Cauchy–Schwarz): (x² + 4y²)(1 + 1/4) ≥ (x + y)² = 100 ⇒ x² + 4y² ≥ 100 × 4/5 = 80, with equality when x/1 = 2y/(1/2), i.e. x = 4y.
Tip: For a sum constraint, the minimum of ax² + by² occurs when ax = by.
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