RagvaloD wrote: Prove for positive $a,b,c$ that $$ (a^2+\frac{b^2}{c^2})(b^2+\frac{c^2}{a^2})(c^2+\frac{a^2}{b^2}) \geq abc (a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})$$ Because: $$(a^2+\frac{b^2}{c^2})(b^2+\frac{c^2}{a^2}) \geq \left(ab+\frac{b}{a}\right)^2=b^2\left(a+\frac{1}{a}\right)^2$$

RagvaloD wrote: Prove for positive $a,b,c$ that $$ (a^2+\frac{b^2}{c^2})(b^2+\frac{c^2}{a^2})(c^2+\frac{a^2}{b^2}) \geq abc (a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})$$ $$\iff$$Prove for positive $a,b,c$ that $$ (a+\frac{b}{c})(b+\frac{c}{a})(c+\frac{a}{b}) \geq (a+1)(b+1)(c+1)$$

sqing wrote: The following inequality is also true. Prove for positive $a,b,c$ that $$ (ab+\frac{c}{a})(bc+\frac{a}{b})(ca+\frac{b}{c}) \geq (ab+1)(bc+1)(ca+1)$$ If $ab=x,bc=y,ac=z$ then we get inequality $(x+\frac{y}{x})(y+\frac{z}{y})(z+\frac{x}{z}) \geq (x+1)(y+1)(z+1)$ $xyz+ \frac{x^2y^2}{yz}+\frac{x^2z^2}{xy}+\frac{y^2z^2}{zx}+\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} +1\geq xyz+xy+yz+xz+x+y+z+1$ Which is true, because $\frac{x^2y^2}{yz}+\frac{x^2z^2}{xy}+\frac{y^2z^2}{zx} \geq xy+yz+xz$ and $\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x} \geq x+y+z$

sqing wrote: Prove for positive $a,b,c$ that $$ (a+\frac{b}{c})(b+\frac{c}{a})(c+\frac{a}{b}) \geq (a+1)(b+1)(c+1)$$ $$\implies$$Prove for positive $a,b,c$ that $$ (ab+\frac{c}{b})(bc+\frac{a}{c})(ca+\frac{b}{a}) \geq (ab+1)(bc+1)(ca+1)$$

We have \begin{align*} &(a^2+\frac{b^2}{c^2})(b^2+\frac{c^2}{a^2})(c^2+\frac{a^2}{b^2}) \geqslant abc (a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c}) \\ \iff &a^4+b^4+c^4+\frac{a^2c^2}{b^2}+\frac{b^2a^2}{c^2}+\frac{c^2b^2}{a^2}\geqslant a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2. \end{align*}This follows from adding up \[a^4+b^4+c^4\geqslant a^2b^2+b^2c^2+c^2a^2\]\begin{align*} &\frac{a^2c^2}{b^2}+\frac{b^2a^2}{c^2}+\frac{c^2b^2}{a^2}\geqslant a^2+b^2+c^2 \\ \iff &a^4b^4+b^4c^4+c^4a^4\geqslant a^2b^2c^2(a^2+b^2+c^2 ). \quad \blacksquare \end{align*}

Very brute force Notice that we have that $\prod_{cyc}a^2+\frac{b^2}{c^2}=a^2b^2c^2+1+\sum_{cyc}\frac{a^2b^2}{c^2}+\sum_{cyc}a^4$ and also $abc\prod_{cyc}a+\frac{1}{a}=a^2b^2c^2+1+\sum_{cyc}a^2b^2+\sum_{cyc}a^2$ Thus the inequality is equivalent to $a^2b^2c^2+1+\sum_{cyc}\frac{a^2b^2}{c^2}+\sum_{cyc}a^4\ge a^2b^2c^2+1+\sum_{cyc}a^2b^2+\sum_{cyc}a^2\Longrightarrow \sum_{cyc}\frac{a^2b^2}{c^2}+\sum_{cyc}a^4\ge\sum_{cyc}a^2b^2+\sum_{cyc}a^2$ also let $x=a^2, y=b^2\text{ and }z=c^2$ So the inequality transforms into $\sum_{cyc}x^2+\sum_{cyc}\frac{xy}{z}\ge\sum_{cyc}xy+\sum_{cyc}x$ Furthermore notice that $\sum_{cyc}x^2\ge\sum_{cyc}xy$ thus all we have left to prove is $\sum_{cyc}\frac{xy}{z}\ge\sum_{cyc}x$ Also notice that $\sum_{cyc}\frac{xy}{z}=\frac{\sum_{cyc}x^2y^2}{xyz}$, thus we have that $\sum_{cyc}x^2y^2\ge xyz\sum_{cyc}x=\sum_{cyc}x^2yz$, however this is clearly true by Muirhead since $(2,2,0)\succ(2,1,1)$ $\blacksquare$.

pretty cool problem we can rewrite the given inequality as √(a^2+b^2/c^2)√(a^2+b^2/c^2)√(b^2+c^2/a^2)√(b^2+c^2/a^2)√(c^2+a^2/b^2)≥√(ab+b/a)^2...(by cauchy) next we write it as √(b^2(a+1/a)^2... = (b(a+1/a)... next we multiply everything and get the result note: i know latex exist its just that my account is new

Keith50 wrote: We have \begin{align*} &(a^2+\frac{b^2}{c^2})(b^2+\frac{c^2}{a^2})(c^2+\frac{a^2}{b^2}) \geqslant abc (a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c}) \\ \iff &a^4+b^4+c^4+\frac{a^2c^2}{b^2}+\frac{b^2a^2}{c^2}+\frac{c^2b^2}{a^2}\geqslant a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2. \end{align*}This follows from adding up \[a^4+b^4+c^4\geqslant a^2b^2+b^2c^2+c^2a^2\]\begin{align*} &\frac{a^2c^2}{b^2}+\frac{b^2a^2}{c^2}+\frac{c^2b^2}{a^2}\geqslant a^2+b^2+c^2 \\ \iff &a^4b^4+b^4c^4+c^4a^4\geqslant a^2b^2c^2(a^2+b^2+c^2 ). \quad \blacksquare \end{align*} How did you get the second line?

Let $a^2=x$;$b^2=y$;$c^2=z$ Then $(xy+z)(yz+x)(xz+y) \geq xyz(x+1)(y+1)(z+1)$ $xyz(x^2+y^2+z^2)+x^2y^2+z^2x^2+y^2z^2 \geq xyz(xy+yz+zx)+xyz(x+y+z)$ $xyz(x^2+y^2+z^2) \geq xyz(xy+yz+zx)$ is obvious $x^2y^2+z^2x^2+y^2z^2 \geq \frac{(xy+yz+zx)^2 }{3} \geq xyz(x+y+z)$ and we are done