Just substitute $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ the desired inequalities is equivalent to $(x^3+y^3+z^3)^2\geq (x^2y+y^2z+z^2x)(xy^2+yz^2+zx^2)$ which is true by rearrangement inequality

This is more directly also equal to the cyclic sum of $ab^2(a-c)^2(a+c)$ is $\ge0$.

just expand! so it remains to rpove that $a^4b^2 +b^4c^2 +c^4a^2+ab^2+bc^2+ca^2\ge 3 +a^2b+b^2c+c^a$ which is direct AM-GM as $a^4b^2+b^4c^2+c^4a^2\ge a^2b+b^2c+c^2a$ and $ab^2+bc^2+ca^2\ge 3abc = 3$ adding yields the result

It's easy to prove using AM-GM that: $$LHS^2 \ge 3(b^2a + c^2b + a^2c)$$ $$LHS^2 \ge 3(a^2b + b^2c + c^2a)$$ $$LHS^2 \ge 9abc$$ Add these 3 inequalities, group all the numbers, take the square root and the inequality is proved. This is the official solution, but on the contest I proved it by expanding it and AM-GM

The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: \[a^2b + b^2c + c^2a +3\ge 2\sqrt{(a+b+c)(ab + bc +ca)}\] Dear Doctor! You are welcome to use BW.

hello, it is equivalent to $9 u^4 v^2+18 u^3 v^3+15 u^2 v^4+x^4 \left(12 u^2+12 u v+12 v^2\right)+x^3 \left(24 u^3+36 u^2 v+60 u v^2+24 v^3\right)+x^2 \left(17 u^4+34 u^3 v+87 u^2 v^2+70 u v^3+17 v^4\right)+x \left(4 u^5+10 u^4 v+48 u^3 v^2+62 u^2 v^3+32 u v^4+6 v^5\right)+6 u v^5+v^6\geq 0$ and this is true. Sonnhard.

It's equivalent also to $2\times2=4$, which is true. Tell me please, Dr Sonnhard Graubner, from which substitution you get your polynomial? Thank you.

Stefan4024 wrote: Let $a,b,c \in \mathbb{R}^{+}$ such that $abc=1$. Prove that: $$a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(ab + bc +ca)}$$ The inequality from Iran 2005: Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]

arqady wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: \[a^2b + b^2c + c^2a +3\ge 2\sqrt{(a+b+c)(ab + bc +ca)}\] Dear Doctor! You are welcome to use BW. The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:\[a^2b + b^2c + c^2a +3\ge a+b+c+ab + bc +ca\]

sqing wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:\[a^2b + b^2c + c^2a +3\ge a+b+c+ab + bc +ca\] It's Schur!

Yeah. Let $a,b,c$ be side lengths of a triangle such that $abc=1$. Prove that$$a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(a^2 + b^2+c^2)}$$

sqing wrote: Let $a,b,c$ be side lengths of a triangle such that $abc=1$. Prove that$$a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(a^2 + b^2+c^2)}$$ $a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(a^2 + b^2+c^2)}\Leftrightarrow\left(a^2b + b^2c + c^2a\right)^2 \ge (a+b+c)(a^2 + b^2+c^2)\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}(a^4b^2-a^4bc-a^3b^2c+a^3c^2b)\geq0\Leftrightarrow\sum_{cyc}a^2(ab-bc)(ab-ac)\geq0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}(ac-bc)^2(b^2+c^2-a^2)\geq0$, which is true because $\sum_{cyc}(b^2+c^2-a^2)=a^2+b^2+c^2>0$ and $\sum_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)=16S^2>0$. Done!

Very nice. Thank arqady.

arqady wrote: It's equivalent also to $2\times2=4$, which is true. Tell me please, Dr Sonnhard Graubner, from which substitution you get your polynomial? Thank you. Hello arqady, i have squared the inequalitiy and then i used BW. Sonnhard.

Dr Sonnhard Graubner wrote: i used BW. Sonnhard. I asked you what is your substitution, which you used for the proof by BW. Thank you.

Nice solution, jeneva! Uhh, rather than: expanding, then use AM-GM, you can also use AM-GM without expanding. Just note \[ \begin{aligned} \sum_{cyc}a^2b&=\sum_{cyc}\left(\tfrac 16 (a^2b+a^2b+c^2a)+\tfrac 16(a^2b+b^2c+a^2b)\right) \\&=\sum_{cyc}\left(\tfrac 16 (a^2b+\tfrac ac+\tfrac cb)+\tfrac 16(a^2b+b^2c+\tfrac ac)\right) \\&\geq \sum_{cyc}\left(\tfrac a2+\tfrac{ab}{2} \right) \\&=\tfrac{a+b+c}{2}+\tfrac{ab+bc+ca}{2} \\&\geq \sqrt{(a+b+c)(ab+bc+ca)}, \end{aligned}\] and that's exactly what had to be proven. Note: I only used cyclic sums because I didn't want to type as much.

Square, rearrange, multiply things by $abc$ and $\frac{1}{abc}$, and it suffices to prove \[\sum_{\text{cyc}} a^4b^2+\sum_{\text{cyc}} a^2b^3c\geq 3a^2b^2c^2+\sum_{\text{cyc}} a^3b^2c\] By AM-GM \[\sum_{\text{cyc}} a^4b^2=\sum_{\text{cyc}}\frac{1}{6}a^4b^2+\frac{1}{6}a^4b^2+\frac{1}{6}a^4b^2+\frac{1}{6}a^4b^2+\frac{1}{6}b^4c^2+\frac{1}{6}c^4a^2\geq \sum_\text{cyc} a^3b^2c,\] and \[\sum_{\text{cyc}} a^2b^3c\geq 3a^2b^2c^2,\] so we are done. $\blacksquare$

The given inequality is equivalent to ${(a^2b + b^2c + c^2a)}^{2}(a+b+c)(ab+bc+ca) \ge {(a+b+c)}^{2}{(ab + bc +ca)}^{2}$. Now, by Cauchy-Schwarz, $(a^2b + b^2c + c^2a)(b+c+a)\geq{(ab + bc +ca)}^{2}$, $(a^2b + b^2c + c^2a)(ab+bc+ca)=(a^2b + b^2c + c^2a)(\frac{1}{b}+\frac{1}{c}+\frac{1}{a})\geq{(a+b+c)}^{2}$

Stefan4024 wrote: Let $a,b,c \in \mathbb{R}^{+}$ such that $abc=1$. Prove that: $$a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(ab + bc +ca)}$$ We have \[(a^2b+b^2c+c^2a)^2-abc(a+b+c)(ab+bc+ca)=\sum{bc^2(a+b)(a-b)^2}\ge{0}\]

sqing wrote: Yeah. Let $a,b,c$ be side lengths of a triangle such that $abc=1$. Prove that$$a^2b + b^2c + c^2a \ge \sqrt{(a+b+c)(a^2 + b^2+c^2)}$$

problem is equivilant to show $(\frac{a}{c}+\frac{b}{a}+\frac{c}{b})^2\geq(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ ie $\sum\frac{a^2}{c^2}+\sum\frac{c}{a}\geq3+\sum\frac{a}{c}$ but by AM GM $\sum\frac{a^2}{c^2}+3+\sum\frac{c}{a}\geq2\sum\frac{a}{c}+3+\sum\frac{c}{a}\geq6+\sum\frac{a}{c}$ done

Squaring both sides and simplifying leaves us to prove that \[ a^4b^2+b^4c^2+c^4a^2+ab^2+bc^2+ca^2 \ge a^2b+b^2c+c^2a+3\]Now we can prove that \[ a^4b^2+b^4c^2+c^4a^2 \ge a^2b+b^2c+c^2a \]by first multiplying the RHS by $abc$. Since $abc=1$ we don't need to change the LHS. Thus we have the inequality \[ a^4b^2+b^4c^2+c^4a^2 \ge a^3b^2c+b^3c^2a+c^3a^2b \]Then by using the weights $\left(\frac{2}{3}, \frac{1}{6}, \frac{1}{6}\right)$ and summing cyclically we get our desired result. So then our original inequality becomes \[ a^4b^2+b^4c^2+c^4a^2+ab^2+bc^2+ca^2 \ge a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \ge a^2b+b^2c+c^2a+3 \]Cancelling out terms, we get \[ ab^2+bc^2+ca^2 \ge 3 \]Which is true by AM-GM. Since all steps are reversible, we are done.

arqady wrote: The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: \[a^2b + b^2c + c^2a +3\ge 2\sqrt{(a+b+c)(ab + bc +ca)}\]Dear Doctor! You are welcome to use BW. It follows from a famous inequality $$x(b+c)+y(c+a)+z(a+b)\geq 2\sqrt{(yz+zx+xy)(bc+ca+ab)},$$which can be proven by Cauchy-Schwarz. Now, let $x=ca,y=ab,z=bc$, we have $$a^{2}b+b^{2}c+c^{2}a+3abc\geq 2\sqrt{abc(a+b+c)(bc+ca+ab)},$$or $$a^{2}b+b^{2}c+c^{2}a+3\geq 2\sqrt{(a+b+c)(bc+ca+ab)}.$$

Good solution.