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StarLex1 wrote: Rewrite as $(a+b+c)(ab+bc+ca-4)>=-3$ AM-GM $(a+b+c)>=3$ AM-HM $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>=3$ $(ab+bc+ca)>=3$ Please, stop writing nonsense. You can not use $a+b+c\ge 3$ when $ab+bc+ca-4$ is negative.

erzhane wrote: StarLex1 wrote: Rewrite as $(a+b+c)(ab+bc+ca-4)>=-3$ AM-GM $(a+b+c)>=3$ AM-HM $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}>=3$ $(ab+bc+ca)>=3$ Please, stop writing nonsense. You can not use $a+b+c\ge 3$ when $ab+bc+ca-4$ is negative. yes, of course

He really need to read Polya books now, especialy 10 advices when solving a problem invented by Polya

Let $(a,b,c) = (x^3,y^3,z^3)$, then we need to prove \[ \sum_{sym} x^6y^3 - 4\sum x^5y^2z^2 + 6x^3y^3z^3 \ge 0 \]It's true because \begin{align*} LHS &= \frac{1}{2}\sum x^2(y-z)^2 \sum x(y^2+yz+z^2)(y-z)^2 + \\ &\quad + xyz \sum x^2(2x^2+y^2+z^2-yz)(y-z)^2 \\ &\ge 0 \end{align*}

mathcyber124 wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$(a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c).$$ pqr method: $p = a + b + c, q = ab + bc + ca, r = abc = 1$. We need to prove that $pq + 3 \ge 4p$. Using $q^2\ge 3pr$, we have $q\ge \sqrt{3pr} = \sqrt{3p}$, and it suffices to prove that $p\sqrt{3p} + 3 \ge 4p$. Since $p\ge 3$, letting $p = 3u^2$ ($u\ge 1$), it suffices to prove that $9u^3 - 12u^2 + 3 \ge 0$ or $3(u - 1)(3u^2 - u - 1) \ge 0$ which is true. We are done.

mathcyber124 wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$(a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c).$$ Note that \begin{align*} ab + bc + ca + \frac{3}{a + b + c} &= 3 \cdot \frac{ab + bc + ca}{3} + \frac{3}{a + b + c} \\ &\ge 4 \sqrt[4]{ \frac{(ab + bc + ca)^3}{9(a + b + c)} } \\ &\ge 4 \sqrt[4]{ \frac{3 \sqrt[3]{a^2 b^2 c^2} \cdot (ab + bc + ca)^2}{9(a + b + c)} } \\ &\ge 4 \sqrt[4]{ \frac{3 \cdot 3abc(a + b + c)}{9(a + b + c)} } \\ &= 4 \end{align*}

This is not the best: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$5(a + b + c)(ab + bc + ca) + 36\ge 27(a + b + c).$$ But there is a very simple proof:

mudok wrote: This is not the best: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$5(a + b + c)(ab + bc + ca) + 36\ge 27(a + b + c).$$ But there is a very simple proof: $$5(a+b+c)(ab+bc+ca)+39 \geq 28(a+b+c).$$

anh1110004 wrote: mudok wrote: This is not the best: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$5(a + b + c)(ab + bc + ca) + 36\ge 27(a + b + c).$$ But there is a very simple proof: $$5(a+b+c)(ab+bc+ca)+39 \geq 28(a+b+c).$$ Stronger: $5(a + b + c)(ab + bc + ca) + \frac{81}{2}\ge \frac{57}{2}(a + b + c)$. We may find the best constant $k > 0$ such that $5(a + b + c)(ab + bc + ca) + 3k - 45\ge k(a + b + c)$ for all $a, b, c > 0$ with $abc = 1$.

$x^4yz+y^4xz+z^4xy+x^3z^3+y^3z^3+x^3y^3+6x^2y^2z^2\ge4(x^3yz^2+y^3zx^2+z^3xy^2)$

lazizbek42 wrote: $x^4yz+y^4xz+z^4xy+x^3z^3+y^3z^3+x^3y^3+6x^2y^2z^2\ge4(x^3yz^2+y^3zx^2+z^3xy^2)$ Any information about it?

$a+b+c=3x^2\geq3 => x\geq1$ $(ab+bc+ac)^2\geq3abc(a+b+c)$ $(ab+bc+ac)\geq3x$ $(a+b+c)(ab+bc+ac)+3\geq9x^3+3$ We prove that $9x^3+3\geq12x^2$ => $(3x^2-x-1)(x-1)\geq0$ This is true $x\geq1$

mathcyber124 wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$(a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c).$$ https://artofproblemsolving.com/community/c6h1933956p13291367

mathcyber124 wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$(a + b + c)(ab + bc + ca) + 3\ge 4(a + b + c).$$ $ab+bc+ca+\frac{3}{a+b+c}\geq \sqrt{3abc(a+b+c)}+\frac{3}{a+b+c}$ $=\sqrt{3}x+\frac{3}{x^2}$ where $x=\sqrt{a+b+c}\geq \sqrt{3}$ by AM-GM $=\frac{(x-\sqrt{3})\{(x-\sqrt{3})(\sqrt{3}x+2)+\sqrt{3}\}}{x^2}+4\geq 4.$ Equality : $x = \sqrt{3}\Longleftrightarrow a = b = c = 1.$