Question

Prove that $0 \le xy+yz+xz - 2xyz \le 7/27, \:\:\:x,y,z \ge 0$ with $x+y+z=1$.

score 861 · Answer 1 · 2019-06-20 09:37:36

861

Views

Prove that $0 \le xy+yz+xz - 2xyz \le 7/27, \:\:\:x,y,z \ge 0$ with $x+y+z=1$.

Published on 20 Jun 2019 - 9:37

score 69 · Answer 2 · 2019-06-21 15:40:53

69

Views

A little confusion about AM-GM proof

Published on 21 Jun 2019 - 15:40

score 266 · Answer 3 · 2026-03-25 11:04:48

266

Views

Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

Published on 25 Mar 2026 - 11:04

score 211 · Answer 4 · 2026-03-25 09:28:46

211

Views

Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ .

Published on 25 Mar 2026 - 9:28

score 168 · Answer 5 · 2026-03-25 22:10:13

168

Views

Find the minimum value of the perimeter

Published on 25 Mar 2026 - 22:10

score 116 · Answer 6 · 2026-03-26 22:18:19

116

Views

Prove that $\sum \frac {xy}{\sqrt{ (x^2+z^2)(y^2+z^2)}}\leq \frac {3}{2} $

Published on 26 Mar 2026 - 22:18

score 78 · Answer 7 · 2026-03-25 06:33:14

78

Views

How can I prove this multilinear inequality?

Published on 25 Mar 2026 - 6:33

score 103 · Answer 8 · 2026-03-26 14:34:48

103

Views

Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that $\frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$ .

Published on 26 Mar 2026 - 14:34

score 193 · Answer 9 · 2026-03-25 07:45:51

193

Views

Given three positive numbers $a,b,c$ so that $abc= 1$. Prove $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a})\leqq\frac{2}{a+b+c-1}$ .

Published on 25 Mar 2026 - 7:45

score 76 · Answer 10 · 2019-07-08 06:28:06

76

Views

Finding root of an equation

Published on 08 Jul 2019 - 6:28

score 46 · Answer 11 · 2019-07-08 16:40:07

46

Views

Why do we need to show equality exists for finding maxima/minima questions?

Published on 08 Jul 2019 - 16:40

score 71 · Answer 12 · 2019-07-08 19:41:23

71

Views

Where is my solution wrong for this AM-GM Maxima question?

Published on 08 Jul 2019 - 19:41

score 84 · Answer 13 · 2026-03-25 08:07:59

84

Views

Inequality in triangle $ABC$ $a=BC$ $b=AC$ $c=AB$

Published on 25 Mar 2026 - 8:07

score 77 · Answer 14 · 2019-07-11 20:47:16

77

Views

If $a,b,c>0$ and $2(a+b+c)=3$ prove $\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4$

Published on 11 Jul 2019 - 20:47

score 141 · Answer 15 · 2026-03-25 07:46:30

141

Views

Given three positive numbers $x,y,z$ so that $x+y+z=\frac{3}{2}$. Prove that $28\,x^2y^2z^2+3(x^2y^2+y^2z^2+z^2x^2)\leqq1$ .

Published on 25 Mar 2026 - 7:46

List Question

Prove that $0 \le xy+yz+xz - 2xyz \le 7/27, \:\:\:x,y,z \ge 0$ with $x+y+z=1$.

A little confusion about AM-GM proof

Given three positive numbers $a,\,b,\,c$ . Prove that $(\!abc+ a+ b+ c\!)^{3}\geqq 8\,abc(\!1+ a\!)(\!1+ b\!)(\!1+ c\!)$ .

Given three non-negative numbers $a, b, c$ so that $a+ b+ c= 3,\,a^{2}+ b^{2}+ c^{2}= 5$. Prove $a^{3}b+ b^{3}c+ c^{3}a\leqq 8$ .

Find the minimum value of the perimeter

Prove that $\sum \frac {xy}{\sqrt{ (x^2+z^2)(y^2+z^2)}}\leq \frac {3}{2} $

How can I prove this multilinear inequality?

Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that $\frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$ .

Given three positive numbers $a,b,c$ so that $abc= 1$. Prove $(a-1+\frac{1}{b})(b-1+\frac{1}{c})(c-1+\frac{1}{a})\leqq\frac{2}{a+b+c-1}$ .

Finding root of an equation

Why do we need to show equality exists for finding maxima/minima questions?

Where is my solution wrong for this AM-GM Maxima question?

Inequality in triangle $ABC$ $a=BC$ $b=AC$ $c=AB$

If $a,b,c>0$ and $2(a+b+c)=3$ prove $\sum_{cyc}\frac{9a}{b^2+2(b+1)}≥4$

Given three positive numbers $x,y,z$ so that $x+y+z=\frac{3}{2}$. Prove that $28\,x^2y^2z^2+3(x^2y^2+y^2z^2+z^2x^2)\leqq1$ .

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