Definite Integration
Integral inequalities using AM-GM and Jensen
MJAT_TS1_P1
Grade 12
Question:
If $f:[0,1]\to(0,\infty)$ is a continuous function such that $\displaystyle\int_0^1 f(x)\,dx = 1$, then which of the following is/are always TRUE?
A) $\left(\displaystyle\int_0^1 f(x)^3\,dx\right)\cdot\left(\displaystyle\int_0^1 f(x)^5\,dx\right) \geq 1$
B) $\displaystyle\int_0^1 \frac{1 + f(x)^2}{2f(x)}\,dx \geq 1$
C) $\left(\displaystyle\int_0^1 f(x)^4\,dx\right)^2 \geq 1$
D) There exists $c\in(0,1)$ such that $f(c) = 1$
Step-by-Step Solution
Key Concept: A: Use Cauchy-Schwarz: $(\int f^3)(\int f^5) \geq (\int f^4)^2 \geq (\int f)^4... $ Also by AM-GM: $f^3 + 1 + 1 \geq 3f$. B: $(1+t^2)/2t \geq 1$ by AM-GM, so the integral $\geq \int_0^1 1\,dx = 1$. D: By IVT — since $f$ is continuous and $\int_0^1 f = 1$, if $f$ were always $> 1$ or always $< 1$ the integral would differ.
A: By Cauchy-Schwarz, $(\int f^3)(\int f^5) \geq (\int f^4)^2$; and by Power Mean $(\int f^4)^{1/4} \geq \int f = 1$, so $(\int f^4)^2 \geq 1$. B: AM-GM gives $\frac{1+f^2}{2f}\geq 1$ pointwise. D: IVT with $\int_0^1 f = 1$. C: False in general (e.g., $f\equiv 1$ gives $= 1$, but $f$ near a constant slightly below 1 gives $<1$). A, B, D are always true.
Correct Answer: ABD