Question

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

score 265 · Answer 1 · 2019-06-23 03:27:02

265

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 23 Jun 2019 - 3:27

score 192 · Answer 2 · 2019-06-27 14:08:53

192

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 27 Jun 2019 - 14:08

score 208 · Answer 3 · 2019-06-29 16:35:29

208

Views

Calculate the maximum value of $|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$ where $a^2 + b^2 + c^2 = 1$.

Published on 29 Jun 2019 - 16:35

score 119 · Answer 4 · 2026-03-25 04:43:24

119

Views

Given numbers $a,b,c\geqq0$ and $-\frac{2}{11}\leqq k\leqq0$. Prove that $(k+1)^{6}(a+b+c)^{2}(\!ab+bc+ca\!)^{2}-81\prod\limits_{sym}(ka+b)\geqq0$ .

Published on 25 Mar 2026 - 4:43

score 326 · Answer 5 · 2019-07-01 09:47:23

326

Views

Given three positive numbers $x,y,z$, prove that $(xyz+x^{2}y+y^{2}z+z^{2}x)^{4}\geqq\frac{256}{27}(x+y+z)^{3}x^{3}y^{3}z^{3}$ .

Published on 01 Jul 2019 - 9:47

score 178 · Answer 6 · 2019-07-04 04:55:53

178

Views

Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$

Published on 04 Jul 2019 - 4:55

score 140 · Answer 7 · 2019-07-14 09:13:14

140

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 14 Jul 2019 - 9:13

score 149 · Answer 8 · 2026-03-22 13:24:35

149

Views

Prove $x+y+z \ge xy+yz+zx$

Published on 22 Mar 2026 - 13:24

score 272 · Answer 9 · 2019-07-15 03:04:42

272

Views

Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .

Published on 15 Jul 2019 - 3:04

score 89 · Answer 10 · 2019-07-17 15:12:45

89

Views

Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$.

Published on 17 Jul 2019 - 15:12

score 125 · Answer 11 · 2026-03-25 03:04:12

125

Views

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$

Published on 25 Mar 2026 - 3:04

score 470 · Answer 12 · 2019-08-05 08:04:55

470

Views

It may be a strengthening form of mean inequality

Published on 05 Aug 2019 - 8:04

score 238 · Answer 13 · 2019-08-08 23:46:45

238

Views

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Published on 08 Aug 2019 - 23:46

score 139 · Answer 14 · 2019-08-11 08:34:21

139

Views

Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$

Published on 11 Aug 2019 - 8:34

score 142 · Answer 15 · 2019-09-17 15:53:12

142

Views

show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$

Published on 17 Sep 2019 - 15:53

List Question

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

Calculate the maximum value of $|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$ where $a^2 + b^2 + c^2 = 1$.

Given numbers $a,b,c\geqq0$ and $-\frac{2}{11}\leqq k\leqq0$. Prove that $(k+1)^{6}(a+b+c)^{2}(\!ab+bc+ca\!)^{2}-81\prod\limits_{sym}(ka+b)\geqq0$ .

Given three positive numbers $x,y,z$, prove that $(xyz+x^{2}y+y^{2}z+z^{2}x)^{4}\geqq\frac{256}{27}(x+y+z)^{3}x^{3}y^{3}z^{3}$ .

Solve for $x,y,z \in \mathbb{R^+}$ ,$x^2+y^2+z^2=xyz+4$ and $xy+yz+zx=2(x+y+z)$

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

Prove $x+y+z \ge xy+yz+zx$

Given three positive numbers $a,b,c$. Prove that $\sum\limits_{cyc}\sqrt{\frac{a+b}{b+1}}\geqq3\sqrt[3]{\frac{4\,abc}{3\,abc+1}}$ .

Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$.

$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$

It may be a strengthening form of mean inequality

Given $x+y+z=3, x,y,z>0 $ how to prove that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x} >= x^2+y^2+z^2$

Maximizing $\frac{a^2+6b+1}{a^2+a}$, where $a=p+q+r=pqr$ and $ab=pq+qr+rp$ for positive reals $p$, $q$, $r$

show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$

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