Search results
Create the page "Zx 2b" on this wiki! See also the search results found.
- ...t to prove that if any numbers <math>x,y,z </math> satisfy <math>xy + yz + zx + 2xyz = 1 </math>, then <math> x+y+z \geq \frac{3}{2} </math>. ...3 < \frac{1}{4} </math>. Adding these inequalities yields <math>xy + yz + zx + 2xyz < 1 </math>, a contradiction.7 KB (1,224 words) - 16:21, 24 October 2022
- <math>xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)</math>. ...with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} =\frac{zx^3}{y}</math>. So the inequality holds with equality if and only if x = y =5 KB (990 words) - 19:37, 19 December 2021
- \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. and solve for <math>m/n = (abc)^2 = a^2b^2c^2</math>15 KB (2,208 words) - 01:25, 1 February 2024
- x=\frac{2b^2-2c^2}{a^2-3b^2-c^2} P=\left(\frac{b^2-c^2}{a^2-2b^2-2c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-4c^2},\frac{a^2-3b^2-c^2}{2a^2-4b^2-47 KB (1,280 words) - 01:22, 6 February 2024
