Dimostrare $\dfrac{a+b}{2}\ge\dfrac{2ab}{a+b}$ per $a,b>0$

Text

Prove that, for every pair of positive numbers $a,b$, the arithmetic mean is not less than the harmonic mean: $\frac{a+b}{2}\ge\frac{2ab}{a+b}.$

Solution

Since $a,b>0$, we have $a+b>0$; compute the difference over a common denominator: $\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)^2-4ab}{2(a+b)}=\frac{(a-b)^2}{2(a+b)}\ge 0.$ The numerator is a square and the denominator is positive, so the ratio is $\ge 0$. Equality iff $a=b$. $\boxed{\frac{a+b}{2}\ge\frac{2ab}{a+b},\qquad =\ \text{iff } a=b.}$