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 Skip Navigation LinksMath Help > Inequalities > Inequality Methods > Young's Inequality

Young's Inequality

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

\[\int_0^af(x)dx+\int_0^bf^{-1}(y)dy\ge ab,\]

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.

Internet references

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.

Related pages in this website

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 Cauchy-Schwarz inequality

The Triangle Inequality

The Quadratic Formula

 


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