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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. Visual proof that (x + y) 2 ≥ 4xy. Taking square roots and dividing by two gives the AM ...
Proof without words of the inequality of arithmetic and geometric means, drawn by CMG Lee. PR is a diameter of a circle centred on O; its radius AO is the arithmetic mean of a and b . Using the geometric mean theorem, right triangle PGR can be split into two similar triangles PQG and GQR; GQ / a = b / GQ, hence GQ = √( ab ), the geometric mean.
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.
There are three inequalities between means to prove. There are various methods to prove the inequalities, including mathematical induction, the Cauchy–Schwarz inequality, Lagrange multipliers, and Jensen's inequality. For several proofs that GM ≤ AM, see Inequality of arithmetic and geometric means.
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1.5 Fifth proof: AM-GM. 1.6 Sixth proof: Titu's lemma. ... Download QR code; Print/export ... We then apply the AM-GM inequality to obtain
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.
English: Four 7 × 9 rectangles packed into a 16 × 16 square, providing a visual proof of the inequality of arithmetic and geometric means and a solution to the two-dimensional analogue of Hoffman's packing puzzle