Definite Integration
Definite — Integral Inequality Proof
Grade 12
Question:
<p>If \(f,g\) are both non-decreasing on \([0,1]\), which inequality holds? [JEE Advanced 2011]</p>
\intfg \geq (\intf)(\intg)
\intfg \leq (\intf)(\intg)
\intfg = (\intf)(\intg)
Cannot compare
Step-by-Step Solution
Key Concept: Chebyshev's sum inequality: if f,g are same-monotone, \int_0^1fg \geq (\int_0^1f)(\int_0^1g).
<div class='solution'>
<p><strong>Chebyshev Integral Inequality:</strong> If $f,g$ are both non-decreasing (or both non-increasing) on $[0,1]$:</p>
<p>$$\int_0^1 f(x)g(x)\,dx\ge\int_0^1 f(x)\,dx\cdot\int_0^1 g(x)\,dx$$</p>
<p><em>Proof idea:</em> Consider $\int_0^1\int_0^1(f(x)-f(y))(g(x)-g(y))dx\,dy\ge 0$. Expanding gives the result.</p>
Correct Answer: A