lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc > = 1$. Prove that $ a^{3} + b^{3} + c^{3} > = ab + bc + ca$. By Muirhead we have: $ a^3+b^3+c^3=\frac{1}{2} \sum_{sym}{a^3} \ge \frac{1}{2}\sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}=\sqrt[3]{abc}(ab+ac+bc) \ge ab+ac+bc$

An easier approach is the following: Chebyshev's inequality implies \[ a^3 +b^3 +c^3 \geq \frac{a+b+c}{3}(a^2 +b^2 +c^2).\] By AM-GM we have $ \frac{a+b+c}{3}\geq (abc)^{\frac{1}{3}}\geq 1,$ and clearly we have $ a^2 +b^2 +c^2 \geq ab+bc+ca$. The result follows.

by letting$a=x^3,b=y^3,c=z^3$ we get$x^9+y^9+z^9\ge x^4y^4z+x^4yz^4+xy^4z^4$ which is trivial by Muirhead

Chbeyshev $\LHS geq\(a+b+c)(a^2+b^2+c^2)

By Hölder, $9(a^{3} + b^{3} + c^{3}) \geq (a+b+c)^3$. By Maclaurin (or trivial expanding), $(a+b+c)^2 \geq 3(ab+bc+ca)$. By AM-GM, $a+b+c \geq 3\sqrt[3]{abc} \geq 3$. So $(a+b+c)^3 \geq 9(ab+bc+ca)$, and we are done.

By AM-GM, $\frac{a^3 + b^3 + 1}{3} \ge ab$ and $a^3 + b^3 \ge 3ab - 1$. Cyclic summing gives $2(a^3 + b^3 + c^3) \ge 3(ab+bc+ac)-3$ and $a^3 + b^3 + c^3 \ge \frac{3}{2} (ab+bc+ac) - \frac{3}{2}$. Again by AM-GM, $\frac{ab+bc+ac}{3} \ge (a^2 b^2 c^2)^\frac{1}{3} \ge 1$ and $\frac{1}{2}(ab+bc+ac) \ge \frac{3}{2}$. Finally, combining both inequalities, we have $a^3 + b^3 + c^3 \ge \frac{1}{2}(ab+bc+ac) + (ab+bc+ac) - \frac{3}{2} \ge ab+bc+ac$ and we win.

if we prove $a^{3}+b^{3}+c^{3}\geqslant a^{2}+b^{2}+c^{2}$ then it will be done. Let's rewrite it as $\frac{a^{4}}{a}+\frac{b^{4}}{b}+\frac{c^{4}}{c}\geqslant a^{2}+b^{2}+c^{2}$, by Cauchy $(a^{2}+b^{2}+c^{2})^2\geqslant (a+b+c)(a^2+b^2+c^2)$ then we have to prove $a^2+b^2+c^2\geqslant a+b+c$ again applying Cauchy $\frac{(a+b+c)^2}{3}\geqslant a+b+c$ and get $a+b+c\geqslant 3$ and by AM-GM $a+b+c\geqslant 3\sqrt[3]{abc}\geqslant 3$, Q.E.D.

\[\Longleftarrow \sum x^9 \ge \sum x^4y^4z \] \[\Longleftrightarrow \sum (x^9+x^9+x^9+x^9+y^9+y^9+y^9+y^9+z^9-9x^4y^4z) \ge 0\] Which is trivial by $ AM-GM \ inequality $

lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] $3(a^3+b^3+c^3) \ge a^3+b^3+c^3+6abc \ge$ $\ge (a+b+c)(ab+bc+ca) \ge 3(ab+bc +ca)$

We know $(abc)^{1/3}(ab+bc+ca)\ge1(ab+bc+ca)$ is true by the given inequality. By Muirhead $2(a^3+b^3+c^3)\ge2(a^{4/3}b^{1/3}c^{1/3}+a^{1/3}b^{4/3}c^{1/3}+a^{1/3}b^{1/3}c^{4/3})$ is true. Dividing both sides of the inequality by 2 tells us that $a^3+b^3+c^3\ge(abc)^{1/3}(ab+bc+ca)\ge1(ab+bc+ca)$ so we are done.

trkac wrote: lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] $3(a^3+b^3+c^3) \ge a^3+b^3+c^3+6abc \ge$ $\ge (a+b+c)(ab+bc+ca) \ge 3(ab+bc +ca)$ Sorry, how to prove it? $$a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $$

Mathskidd wrote: trkac wrote: lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] $3(a^3+b^3+c^3) \ge a^3+b^3+c^3+6abc \ge$ $\ge (a+b+c)(ab+bc+ca) \ge 3(ab+bc +ca)$ Sorry, how to prove it? $a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $ Simplifying it we see it is equivalent to $a^3+b^3+c^3+3abc \ge \sum a^2b$. Note that if we prove $3abc\ge a^3+b^3+c^3$ then we will be done as it will prove that $a^3+b^3+c^3+3abc \ge 2(a^3+b^3+c^3) \ge \sum a^2b$. Not sure how to do this give restriction $abc\ge1$, for a second I got the inequality was not true oops

@anandlyer Are you sure $3abc\ge a^3+b^3+c^3$ But AM_GM $ a^3+b^3+c^3\ge 3abc$

Mathskidd wrote: Sorry, how to prove it? $$a^3+b^3+c^3+6abc \ge (a+b+c)(ab+bc+ca) $$ It's just Schur. We can prove it by very many ways. For example: Since this inequality is symmetrical, we can assume $a\geq b\geq c$. Hence, $a^3+b^3+c^3+6abc - (a+b+c)(ab+bc+ca)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$ $=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=(a-b)^2(a+b-c)\geq0$.

lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] We have \[\sum{a^3}-\sum{bc}=\frac{1}{3}\sum{(2a+1)(a-1)^2}+\frac{3+a+b+c}{6}\sum{(b-c)^2}+(abc-1)\ge{0}\]

By AM-GM, $a+b+c \ge 3 \sqrt[3]{abc} \ge 3$, so $a^2+b^2+c^2 \ge \frac{(a+b+c)^2}{3} \ge a+b+c$. Because $a^2+b^2+c^2 \ge ab+bc+ca$, it follows by Cauchy-Schwarz that $a^3+b^3+c^3 \ge \frac{(a^2+b^2+c^2)^2}{a+b+c} \ge ab+bc+ca$.

lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] The following inequality is also true : Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ 2( a^{3} + b^{3} + c^{3} )+3 \geq 3( ab + bc + ca ).\]

abdelkrim wrote: The following inequality is also true : Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ 2( a^{3} + b^{3} + c^{3} )+3 \geq 3( ab + bc + ca ).\] $\sum_{cyc}(a^3+a^3+1)\ge \sum_{cyc}(3a^2)\ge \sum_{cyc}(3ab)$

By the Power Mean and AM-GM Inequalities, $(\frac{a^3+b^3+c^3}{3})^\frac{1}{3} \geq (\frac{a^2+b^2+c^2}{3})^\frac{1}{2} \implies a^3+b^3+c^3 \geq 3(\frac{a^2+b^2+c^2}{3})^\frac{3}{2}$ $= 3(\frac{a^2+b^2+c^2}{3})^\frac{1}{2} (\frac{a^2+b^2+c^2}{3}) \geq (abc)^\frac{1}{3} (ab+bc+ca) \geq ab+bc+ca$.

We can even use sophie Germain's identity

Let $ a,b,c$ be positive numbers, satisfying $ abc= 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq \frac{9}{8}(a+ b+ c)^2-\frac{57}{8}.\]

sqing wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc= 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq \frac{9}{8}(a+ b+ c)^2-\frac{57}{8}.\] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Thus, we need to prove that $f(v^2)\geq0,$ where $$f(v^2)=9u^3-9uv^2+w^3-\frac{27}{8}u^2w+\frac{19}{8}w^3.$$But we see that $f$ decreases, which says that it's enough to prove our inequality for the maximal value of $v^2$, which happens for equality case of two variables. Let $b=a$. Hence, $c=\frac{1}{a^2}$ and we need to prove that $$2a^3+\frac{1}{a^6}\geq\frac{9}{8}\left(2a+\frac{1}{a^2}\right)^2-\frac{57}{8}$$or $$(a-1)^2(16a^7-4a^6-24a^5+13a^4+14a^3+15a^2+16a+9)\geq0,$$which is obvious. The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$a^3+b^3+c^3\geq\frac{4}{3}(a+b+c)^2-9.$$

arqady wrote: sqing wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc= 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq \frac{9}{8}(a+ b+ c)^2-\frac{57}{8}.\] Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$. Thus, we need to prove that $f(v^2)\geq0,$ where $$f(v^2)=9u^3-9uv^2+w^3-\frac{27}{8}u^2w+\frac{19}{8}w^3.$$But we see that $f$ decreases, which says that it's enough to prove our inequality for the maximal value of $v^2$, which happens for equality case of two variables. Let $b=a$. Hence, $c=\frac{1}{a^2}$ and we need to prove that $$2a^3+\frac{1}{a^6}\geq\frac{9}{8}\left(2a+\frac{1}{a^2}\right)^2-\frac{57}{8}$$or $$(a-1)^2(16a^7-4a^6-24a^5+13a^4+14a^3+15a^2+16a+9)\geq0,$$which is obvious. The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$a^3+b^3+c^3\geq\frac{4}{3}(a+b+c)^2-9.$$ Thanks.

My Solution: Just note,that due to Muirhead's inequality,we have: $a^{3}+b^{3}+c^{3}\geq\sqrt[3]{abc}(ab+bc+ca)$ because it's equivalent to $a^{3}+b^{3}+c^{3}\geq\sqrt[3]{a^{4}b^{4}c}+\sqrt[3]{b^{4}c^{4}a}+\sqrt[3]{c^{4}a^{4}b}$ and we know,that$(3,0,0)\succ(\frac{4}{3},\frac{4}{3},\frac{1}{3})$ And since $abc\geq1$,we have: $LHS\geq\sqrt[3]{abc}(ab+bc+ca)\geq ab+bc+ca$ $Q.E.D$

$a^3+b^3+1 \geq 3ab$ $3(a^3+b^3+c^3) \geq 2(a^3+b^3+c^3)+3$ Tugadi.

Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$. Prove that $$a^3+ b^2+ c^2 \geq a^2 +b+c$$$$a^3+ b^2+ c \geq a^2 +b+1$$

sqing wrote: Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$. Prove that $$a^3+ b^2+ c^2 \geq a^2 +b+c$$

Attachments:

Homogenize by multiplying $\sqrt[3]{abc}$, then trivial by Muirhead

sqing wrote: Let $ a,b,c$ be positive numbers satisfying $ abc\geq 1$. Prove that $$a^3+ b^2+ c \geq a^2 +b+1$$ $$\frac{a^2+1}{2}+b+c\geq a+b+c\geq 3$$$$ \frac{3}{2}a^2+2b+c- \frac{3}{2}\geq a^2 +b+1$$$$a^3+ b^2+ c\geq \frac{3}{2}a^2+2b+c- \frac{3}{2}\geq a^2 +b+1$$

Thanks.

from $Power$ $Mean$ $Inequality$ , $(a^3+b^3+c^3) \ge 3abc$ and $(a^2+b^2+c^2) \ge (ab+bc+ac)$ $3(a^3+b^3+c^3)^2 \ge (a^2+b^2+c^2)^3$ $3(a^3+b^3+c^3)^2 \ge (a^2+b^2+c^2)^3 \ge 3(ab+bc+ac)^3$ $\frac{(a^3+b^3+c^3)^3}{abc} \ge 3(a^3+b^3+c^3)^2 \ge 3abc(ab+bc+ac)^3$ $\implies$ $ a^3+b^3+c^3 \ge \sqrt[3]{abc}(ab+ac+bc) \ge ab +bc +ca $ PROVED BY : O.Y.SH.

lasha wrote: Let $ a,b,c$ be positive numbers, satisfying $ abc\geq 1$. Prove that \[ a^{3} + b^{3} + c^{3} \geq ab + bc + ca.\] $\Leftrightarrow \sum_{sym}{a^3}\ge \sum_{sym}{ab}$ As: $abc\ge 1 \Rightarrow \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}\ge\sum_{sym}{ab}$ It is enough to prove that: $\sum_{sym}{a^3}\ge \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}$ As: $3\ge \frac{4}{3}$ $3\ge \frac{8}{3}$ $3=3$ By Muirhead: $\sum_{sym}{a^3}\ge \sum_{sym}{a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}}_\blacksquare$