Table of Contents
(a) For any two positive numbers. A.M. ≥ G.M. ≥ H.M.
Proof : let x1 and x2 be two the two positive numbers. therefore their,
![\begin{array}{l} A.M = \frac{{x}_{1} + {x}_{2}}{2}; G.M. = \sqrt{{x}_{1} {x}_{2}} and H.M. = \frac{2}{\frac{1}{{x}_{1}} + \frac{1}{{x}_{2}}}\\ \\ now -we -know , (\sqrt{{x}_{1}} - \sqrt{{x}_{2}}{)}^{2} \geq 0 [squere-quantity is -nonnegetive]\\ \\ or, {x}_{1} + {x}_{2} - 2\sqrt{{x}_{1} {x}_{2}} \geq 0 \\ \\ or, {x}_{1} + {x}_{2} \geq 2\sqrt{{x}_{1} {x}_{2}}\\ \\ or, \frac{{x}_{1} + {x}_{2}}{2} \geq \sqrt{{x}_{1} {x}_{2}}\\ \\ \therefore A.M. \geq G.M. ............ (1)\\ \\ Again, (\frac{1}{\sqrt{{x}_{1}}} - \frac{1}{\sqrt{{x}_{2}}}{)}^{2} \geq 0 \\ \\ or, \frac{1}{{x}_{1}} + \frac{1}{{x}_{2}} \geq \frac{2}{\sqrt{{x}_{1} {x}_{2}}}\\ \\ or, \sqrt{{x}_{1} {x}_{2}} \geq \frac{2}{\frac{1}{{x}_{1}} + \frac{1}{{x}_{2}}}\\ \\ \therefore G.M. \geq H.M. ............ (2)\\ \\ From (1) and (2) : A.M. \geq G.M. \geq H.M.\\ \\ \\ \end{array}](https://raynoc.com/wp-content/ql-cache/quicklatex.com-767a18268f0c07c1af33b375563a0f78_l3.png "Rendered by QuickLaTeX.com")
(b) for pair of positive number A.M. × H.M = G.M.2
Proof : let positive number be x1, x2.
therefore,
Example: The A.M. and G.M. of two number are respectivly 5 and 4. find their H.M.?
we know ,
A.M. × H.M = G.M.2
therefore, 
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