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- ...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) - 15:21, 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 =3 KB (622 words) - 06:40, 22 May 2024
- \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) - 00:25, 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 (1,290 words) - 15:16, 15 September 2024
- ...AB = 3, AD \cdot BC = 3 \cdot 1 = 3 = (a+c)(b+ d) = (3 - b + 1 - b) \cdot 2b.</cmath> <cmath>\left\{\begin{array}{l} (x^2 + xy + y^2) (y^2 + yz + z^2)(z^2 + zx + x^2) = xyz,\\(x^4 + x^2\cdot y^2 + y^4) (y^4 + y^2 \cdot z^2 + z^4)(z^4 +48 KB (8,214 words) - 07:59, 30 July 2024
