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Let f be a real-valued, continuous, and strictly increasing function on [0,c] with c > 0. If f(0)=0, a in [0,c], and b in [0,f(c)], then
where f-1 is the inverse function of f. Equality holds iff b=f(a). To prove this, draw the graph of f(x), and treat each integral as the area bounded by the x and y axes and the function. Clearly, all of the rectangle bounded by the axes, a, and b is included in the sum of these areas.
. . . . . . a picture will be added to this web page in due course.
Taking the particular function f(x)=xp-1 gives the special case
(ap)/p+((p-1)/p)bp/(p-1) ≥ ab,
which is often written in the symmetric form
(ap)/p+(bq)/q ≥ ab,
where a,b ≥ 0, p>1, and 1/p+1/q=1.
Mathworld: Young's Inequality
Wikipedia: Young's Inequality, which treats only the particular case (ap)/p+(bq)/q ≥ ab, and notes that Young's inequality is used in the proof of the H�lder inequality.
The Nondecreasing Sequence Two puzzle whose most elegant solution relies on Young's Inequality.
The AM-GM Inequality: the Arithmetic Mean of positive numbers is always greater than the Geometric Mean. This is proved using Jensen's Inequality.
The webmaster and author of this Math Help site is Graeme McRae.
![\[\int_0^af(x)dx+\int_0^bf^{-1}(y)dy\ge ab,\]](IneqYoungsInequality1.gif){kind=small}