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[[File:AM_GM_inequality_visual_proof.svg|thumb|{{{1|right}}}|[[Proof without words]] of the [[inequality of arithmetic and geometric means]]:<br style="margin-bottom:1ex;"/><math>PR</math> is the diameter of a circle centered on <math>O</math>; its radius <math>AO</math> is the [[arithmetic mean]] of <math>a</math> and <math>b</math>. {{nowrap|Using the [[geometric mean theorem]],}} {{nowrap|triangle <math>PGR</math>'s}} [[altitude_(triangle)|altitude]] <math>GQ</math> is the [[geometric mean]]. For any ratio {{nowrap|<math>a:b</math>,}} {{nowrap|<math>AO\ge GQ</math>.}}]]<noinclude>
[[File:AM_GM_inequality_visual_proof.svg|thumb|{{{1|right}}}|[[Proof without words]] of the {{nowrap|[[AM–GM inequality]]:}}<br style="margin-bottom:1ex;"/>PR is the diameter of a circle centered on O; its radius AO is the [[arithmetic mean]] of ''a'' and ''b''. Using the [[geometric mean theorem]], triangle PGR's [[altitude_(triangle)|altitude]] GQ is the [[geometric mean]]. For any ratio {{nowrap|''a'':''b'',}} {{nowrap|AO &ge; GQ.}}]]<noinclude>
[[Category:Geometry templates]]
[[Category:Geometry templates]]
[[Category:Mathematics image insertion templates]]
[[Category:Mathematics image insertion templates]]

Latest revision as of 10:20, 1 March 2024

Proof without words of the AM–GM inequality:
PR is the diameter of a circle centered on O; its radius AO is the arithmetic mean of a and b. Using the geometric mean theorem, triangle PGR's altitude GQ is the geometric mean. For any ratio a:b, AO ≥ GQ.