Bunya-Calculus!

Calculus Level 4

If f ( x ) + g ( x ) + h ( x ) = 2 f(x)+g(x)+h(x)=2 for all x ∈ R x \in \mathbb{R} , find the minimum value of the following expression

∫ 0 3 / 4 ( f ( x ) 2 + g ( x ) 2 + h ( x ) 2 ) d x \int_{0}^{3/4}\left ( f(x)^{2}+g(x)^2+h(x)^2 \right )dx


The answer is 1.

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Adrian Castro by Adrian Castro , 23, USA

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1 solution

Adrian Castro
Dec 31, 2015

By the Cauchy-Bunyakovsky-Schwarz Inequality we have

( 1 2 + 1 2 + 1 2 ) ( f ( x ) 2 + g ( x ) 2 + h ( x ) 2 ) ≥ ( 1 ⋅ f ( x ) + 1 ⋅ g ( x ) + 1 ⋅ h ( x ) ) 2 (1^2+1^2+1^2)(f(x)^2+g(x)^2+h(x)^2)\geq (1\cdot f(x)+1\cdot g(x)+1\cdot h(x))^2

⇒ ( f ( x ) 2 + g ( x ) 2 + h ( x ) 2 ) ≥ 4 3 \Rightarrow (f(x)^2+g(x)^2+h(x)^2)\geq \frac{4}{3}

Therefore, the minimum value of the expression is given by

∫ 0 3 / 4 ( 4 3 ) d x = 1 \int_{0}^{3/4}(\frac{4}{3})dx=1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

No Bunya here.... just Cauchy

Otto Bretscher - 5 years, 5 months ago

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Took me a few moments of reading your reply to understand the joke haha. It is a joke right?

Adrian Castro - 5 years, 5 months ago

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I'm pointing out, in a light-hearted manner, that you are just using Cauchy's humble inequality (the finite-dimensional case), not Bunyakovsky's (or Schwarz's) fancier version

Otto Bretscher - 5 years, 5 months ago

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