enow.com Web Search

Search results

  1. Results from the WOW.Com Content Network
  2. AM–GM inequality - Wikipedia

    en.wikipedia.org/wiki/AM–GM_inequality

    Proof without words of the AMGM 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 ...

  3. QM-AM-GM-HM inequalities - Wikipedia

    en.wikipedia.org/wiki/QM-AM-GM-HM_Inequalities

    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 GMAM, see Inequality of arithmetic and geometric means.

  4. Nesbitt's inequality - Wikipedia

    en.wikipedia.org/wiki/Nesbitt's_inequality

    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

  5. Geometric mean theorem - Wikipedia

    en.wikipedia.org/wiki/Geometric_mean_theorem

    Another application of this theorem provides a geometrical proof of the AMGM 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.

  6. Geometric mean - Wikipedia

    en.wikipedia.org/wiki/Geometric_mean

    Proof without words of the AMGM 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.

  7. Template:AM GM inequality visual proof.svg - Wikipedia

    en.wikipedia.org/wiki/Template:AM_GM_inequality...

    Proof without words of the AMGM 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.

  8. Talk:AM–GM inequality - Wikipedia

    en.wikipedia.org/wiki/Talk:AM–GM_inequality

    Yes, I think that's it. There are at least five different proofs of this inequality out there, at least I know five. I am not sure if this particular one can be attributed to whom. I am not even sure if we are talking about the same Newman! —Preceding unsigned comment added by 128.226.170.133 (talk • contribs)

  9. File:QM AM GM HM inequality visual proof.svg - Wikipedia

    en.wikipedia.org/wiki/File:QM_AM_GM_HM...

    You are free: to share – to copy, distribute and transmit the work; to remix – to adapt the work; Under the following conditions: attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made.