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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 ...
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.
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.
Another application of this theorem provides a geometrical proof of the AM–GM inequality in the case of two numbers. For the numbers p and q one constructs a half circle with diameter p + q. Now the altitude represents the geometric mean and the radius the arithmetic mean of the two numbers.
Bernoulli's inequality can be proved for case 2, in which is a non-negative integer and , using mathematical induction in the following form: we prove the inequality for {,}, from validity for some r we deduce validity for +.
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.
A closed-form formula for the characteristic function ... This is due to the AM–GM inequality and is a consequence of the logarithm being a concave function. In fact,
This is a generalization of the inequality of arithmetic and geometric means and a special case of an inequality for generalized means. The proof follows from the arithmetic–geometric mean inequality , AM ≤ max {\displaystyle \operatorname {AM} \leq \max } , and reciprocal duality ( min {\displaystyle \min } and max {\displaystyle \max ...