The size order of the six usual means of two positive numbers ($a$ and $b$ ) is from the least to the greatest one

1. harmonic mean,
2. geometric mean,
3. Heronian mean,
4. arithmetic mean,
5. quadratic mean,
6. contraharmonic mean,

Proof. If $x^2-y^2 \geqq 0$ for nonnegative $x$ and $y$ , then $x \geqq y$ .

``$1\leqq2$ '':
$\displaystyle\left(\sqrt{ab}\right)^2-\left(\frac{a\!+\!b}{2}\right)^2 = ab\!-\!\frac{4a^2b^2}{(a\!+\!b)^2} = ab\left(1\!-\!\frac{4ab}{(a\!+\!b)^2}\right) = ab\cdot\frac{(a\!+\!b)^2-4ab}{(a\!+\!b)^2} = \frac{ab(a\!-\!b)^2}{(a+b)^2} \geqq 0$

``$2\leqq3$ '' and ``$3\leqq4$ '': proven in Heronian mean is between geometric and arithmetic mean

``$4\leqq5$ '':
$\displaystyle\left(\sqrt{\frac{a^2\!+\!b^2}{2}}\right)^2-\left(\frac{a\!+\!b}{2}\right)^2 = \frac{2a^2\!+\!2b^2\!-\!a^2\!-\!2ab\!-\!b^2}{4} = \left(\frac{a\!-\!b}{2}\right)^2 \geqq 0$

``$5\leqq6$ '':
$\displaystyle\left(\frac{a^2\!+\!b^2}{a\!+\!b}\right)^2-\left(\sqrt{\frac{a^2\!+\!b^2}{2}}\right)^2 = \frac{2(a^2\!+\!b^2)^2-(a^2\!+\!b^2)(a\!+\!b)^2}{2(a\!+\!b)^2} = \frac{(a^2\!+\!b^2)(2a^2\!+\!2b^2\!-\!a^2\!-\!2ab\!-\!b^2)}{2(a\!+\!b)^2}\\ = \frac{(a^2\!+\!b^2)(a\!-\!b)^2}{2(a\!+\!b)^2} \geqq 0$