v_Enhance wrote: Prove that for all positive reals $a,b,c$, \[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1} \] $\sum_{cyc}\frac{1}{a+\frac{1}{b}+1}-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=$ $=\sum_{cyc}\frac{1}{a+\frac{1}{b}+1}-1+1-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=$ $=-\frac{(abc-1)^2}{\prod\limits_{cyc}(ab+b+1)}+\frac{\left(\sqrt[3]{abc}-1\right)^2}{\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1}=$ $=\frac{\left(\sqrt[3]{abc}-1\right)^2\left(\prod\limits_{cyc}(ab+b+1)-\left(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1\right)^3\right)}{\left(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1\right)\prod\limits_{cyc}(ab+b+1)}\geq0$, where the last inequality is true by Holder.

Any more solution ???

Using substitution: $a=r\frac{x}{y}, \; b=r\frac{y}{z}, \; c=r\frac{z}{x}$ for $r>0$, inequality is equivalent to: $\sum \frac{y}{r^2x+ry+z} \ge \frac{3}{r^2+r+1}$ Using Cauchy-Schwarz, we have: $\sum \frac{y}{r^2x+ry+z} \ge \frac{(x+y+z)^2}{(r^2+1)\sum xy+r\sum x^2}$ Hence, it remains to prove: $\frac{(x+y+z)^2}{(r^2+1)\sum xy+r\sum x^2}\ge \frac{3}{r^2+r+1}$ What is equivalent to: $(r-1)^2\left( \sum a^2-\sum ab \right) \ge 0$ and it is obviously true.$\blacksquare$

MathUniverse wrote: Using substitution: $a=r\frac{x}{y}, \; b=r\frac{y}{z}, \; c=r\frac{z}{x}$ for $r>0$, Sorry , but why we can use this substitution ???

War-Hammer wrote: MathUniverse wrote: Using substitution: $a=r\frac{x}{y}, \; b=r\frac{y}{z}, \; c=r\frac{z}{x}$ for $r>0$, Sorry , but why we can use this substitution ??? Because if $a,b,c>0$, then $abc=r^3$ for some $r>0$. Therefore, we can make substitution: $a'=\frac{a}{r}$ and so on, so we get $a'b'c'=1$. Then we make standard substitution: $a'=\frac{x}{y}, \; b'=\frac{y}{z}, \; c'=\frac{z}{x}$. Combining two substitutions, we get written one. I hope it's clear now.

arqady wrote: v_Enhance wrote: Prove that for all positive reals $a,b,c$, \[\frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1}\ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1} \] $\sum_{cyc}\frac{1}{a+\frac{1}{b}+1}-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=$ $=\sum_{cyc}\frac{1}{a+\frac{1}{b}+1}-1+1-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=$ $=\frac{(abc-1)^2}{\prod\limits_{cyc}(ab+b+1)}+\frac{\left(\sqrt[3]{abc}-1\right)^2}{\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1}\geq0$. sorry but I think that the sign of the first number in the last relation is reverse...because if you're right,then $ LHS $ is always at least $ 1 $ which is not true ($ a=b=c $).

Thank you, anonymouslonely! I have fixed my post.

Nice problem. Let $k^3=abc$ and let $a=\frac{ky}{z}, b=\frac{kz}{x}, c=\frac{kx}{y}$. The inequality becomes $\sum_{cyc} \frac{z}{x+k^2y+kz} \geq \frac{3}{k^2+k+1}$. This follows from Cauchy as $\sum_{cyc} \frac{z}{x+k^2y+kz} \geq \frac{(x+y+z)^2}{k(x^2+y^2+z^2)+(k^2+1)(xy+yz+zx)}$, and this last expression is greater than $\frac{3}{k^2+k+1}$ if and only if $(k-1)^2(x^2+y^2+z^2-xy-yz-zx) \geq 0$, upon expansion.

quick and easy! solution= let $a=\frac{kx}{y},b=\frac{ky}{z},c=\frac{kz}{x}$ than the inequality is equivalent to $\sum \frac{y}{k^2x+ky+z}\ge \frac{3}{k^2+k+1}$ which by cauchy is equivalent to $\sum \frac{y}{k^2x+ky+z}\ge \frac{(x+y+z)^2}{\sum (k^2+1)(xy)+\sum k(x^2)}\ge \frac{3}{k^2+k+1}$ as $\frac{(x+y+z)^2}{\sum (k^2+1)(xy)+\sum k(x^2)}\ge \frac{3}{k^2+k+1}\Longrightarrow (k-1)^2(\sum x^2-\sum xy)\ge 0$ as $\sum x^2\ge \sum xy$ so we are done

WOW!!! how could you think of the substitution? is there any reason?

The inequality is \[-\frac{b}{ab + b + 1} - \frac{c}{bc + c + 1} - \frac{a}{ca + a + 1} \le -\frac{3}{\sqrt[3]{abc} + \frac{1}{\sqrt[3]{abc}} + 1}.\]Adding one to both sides and simplifying: \begin{eqnarray*} \frac{(abc - 1)^2}{(ab + b + 1)(bc + c + 1)(ca + a + 1)} & \le & 1 - \frac{3}{\sqrt[3]{abc} + \frac{1}{\sqrt[3]{abc}} + 1}\\ & = & 1 - \frac{3\sqrt[3]{abc}}{\left(\sqrt[3]{abc}\right)^2 + \sqrt[3]{abc} + 1}\\ & = & \frac{\left(\sqrt[3]{abc} - 1\right)^2}{\left(\sqrt[3]{abc}\right)^2 + \sqrt[3]{abc} + 1}\\ & = & \frac{\left(\sqrt[3]{abc} - 1\right)^3}{abc - 1}, \end{eqnarray*}where we assume $abc \neq 1$ (otherwise equality holds). Clearing denominators and noting that $\left(\sqrt[3]{abc} - 1\right)^3(abc - 1)$ is positive, it suffices to show \[(ab + b + 1)(bc + c + 1)(ca + a + 1) \ge \frac{(abc - 1)^3}{\left(\sqrt[3]{abc} - 1\right)^3} = \left(\left(\sqrt[3]{abc}\right)^2 + \sqrt[3]{abc} + 1\right)^3.\]However, this is trivial by Holder's.

$(a+\frac{1}{b}+1)(a+c^2+b^3) \ge (\sum a)^2 \implies \frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1} \ge \frac{\sum a^3 + \sum a^2 + \sum a}{(\sum a)^2}$ then its easily to see that $\frac{\sum a^3 + \sum a^2 + \sum a}{(\sum a)^2} \ge 1 \ge \frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}$

Let us define \[ u=a+\frac{1}{b}+1,\quad v=b+\frac{1}{c}+1,\quad w=c+\frac{1}{a}+1. \] Lemma: We have $$uvw-uv-vw-wu=abc+\frac{1}{abc}-2=\left(\sqrt{abc}-\frac{1}{\sqrt{abc}}\right)^2.$$Proof: Bashing yields, $$uvw=4+2\sum_{cyc}a+2\sum_{cyc}\frac{1}{a}+\sum_{cyc}bc+\sum_{cyc}\frac{1}{bc}+\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)+abc+\frac{1}{abc}.$$We also compute, $$uv+vw+wu=6+2\sum_{cyc}a+2\sum_{cyc}\frac{1}{a}+\sum_{cyc}bc+\sum_{cyc}\frac{1}{bc}+\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right).$$It's easy to see the difference between the two sums matches with the RHS of the lemma. $\square$ Note that the LHS of the inequality is \begin{align*}\frac{1}{u}+\frac{1}{v}+\frac{1}{w}=\frac{uv+vw+wu}{uvw}=\frac{uvw-\left(\sqrt{abc}-\frac{1}{\sqrt{abc}}\right)^2}{uvw}=1- \frac{\left(\sqrt{abc}-\frac{1}{\sqrt{abc}}\right)^2}{uvw}\end{align*}Hence, it's enough to prove that $$1-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}\ge \frac{\left(\sqrt{abc}-\frac{1}{\sqrt{abc}}\right)^2}{uvw}=\frac{\left((\sqrt[6]{abc})^3-\frac{1}{(\sqrt[6]{abc})^3}\right)^2}{uvw}.$$Write that LHS as $$1-\frac{3}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=\frac{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}-2}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}=\frac{\left(\sqrt[6]{abc}-\frac{1}{\sqrt[6]{abc}}\right)^2}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}$$Clearly this inequality would hold if $abc=1$. Assume this isn't the case. Then, we wish to show $$\frac{\left(\sqrt[6]{abc}-\frac{1}{\sqrt[6]{abc}}\right)^2}{\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1}\ge \frac{\left((\sqrt[6]{abc})^3-\frac{1}{(\sqrt[6]{abc})^3}\right)^2}{uvw},$$or, \begin{align*} uvw &\ge \left(\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1\right)\frac{\left((\sqrt[6]{abc})^3-\frac{1}{(\sqrt[6]{abc})^3}\right)^2}{\left(\sqrt[6]{abc}-\frac{1}{\sqrt[6]{abc}}\right)^2}\\ &=\left(\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1\right)\left((\sqrt[6]{abc})^2+1+\frac{1}{(\sqrt[6]{abc})^2}\right)^2\\ &= \left(\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1\right)^3. \end{align*}But the last inequality follows by generalized Holder, so we are done.

H.HAFEZI2000 wrote: $(a+\frac{1}{b}+1)(a+c^2+b^3) \ge (\sum a)^2 \implies \frac{1}{a+\frac{1}{b}+1}+\frac{1}{b+\frac{1}{c}+1}+\frac{1}{c+\frac{1}{a}+1} \ge \frac{\sum a^3 + \sum a^2 + \sum a}{(\sum a)^2}$ Your second inequality is wrong. In fact the reverse holds because $$\frac{1}{a+\frac{1}{b}+1}\le \frac{a+c^2+b^3}{(\sum a)^2}.$$

Let \(a=\sqrt[3]{abc}\cdot\frac xy\), \(b=\sqrt[3]{abc}\cdot\frac yz\), \(c=\sqrt[3]{abc}\cdot\frac zx\), so the desired inequality is equivalent to \[ \sum_\mathrm{cyc}\frac y{\sqrt[3]{abc}\cdot x+y+\frac1{\sqrt[3]{abc}}\cdot z} \stackrel?\ge\frac3{\sqrt[3]{abc}+\frac1{\sqrt[3]{abc}}+1}. \]Define the quantities \begin{align*} A&=\sqrt[3]{abc}\cdot x+y+\tfrac1{\sqrt[3]{abc}}\cdot z\\ B&=\tfrac1{\sqrt[3]{abc}}\cdot x+\sqrt[3]{abc}\cdot y+z\\ C&=x+\tfrac1{\sqrt[3]{abc}}\cdot y+\sqrt[3]{abc}\cdot z. \end{align*}Hence \(A+B+C=(x+y+z)\left(\sqrt[3]{abc}+\frac1{\sqrt[3]{abc}}+1\right)\), so the inequality rewrites as \[ \sum_\mathrm{cyc}\frac yA\stackrel?\ge\frac{3(x+y+z)}{A+B+C}\iff \sum_\mathrm{cyc}\frac y{x+y+z}\cdot\frac1A\ge\frac1{(A+B+C)/3}, \]which is true by Jensen.

Let, $$x = a + \frac1b , y = b + \frac1c , z = c + \frac1a , t = \sqrt[3]{abc}$$Observe that $$xyz - (x+y+z) = t^3 - 3t$$Hence, \begin{align*} & \qquad ~~~ \frac{1}{x + 1} + \frac{1}{y+1} + \frac{1}{z+1} \ge \frac{3}{t+1} \\ & \iff \frac{(xy + yz + zx) + 2(x+y+z) + 3}{(x+1)(y+1)(z+1)} \ge \frac{3}{t+1} \\ &\iff 1 + \frac{xyz - (x+y+z) + 2}{(x+1)(y+1)(z+1)} \ge \frac{3}{t+1} \\ &\iff \frac{t^3 - 3t + 2}{(x+1)(y+1)(z+1)} \ge \frac{-t + 2}{t+1} \\ &\iff \frac{(2-t)(t+1)^2}{(x+1)(y+1)(z+1)} \ge \frac{2-t}{t+1} \\ &\iff (t-2)(x+1)(y+1)(z+1) \ge (t-2)(t+1)^3 \end{align*}Since $t \ge 2$, so we may cancel the $t-2$ factor (otherwise we are just done), which gives $$ \left( a + \frac1b + 1 \right) \left( b + \frac1c + 1 \right) \left( c + \frac1a + 1 \right) \ge \left(\sqrt[3]{abc} + \sqrt[3]{\frac{1}{abc}} + 1 \right)^3 $$And that just follows by Holder's. $\blacksquare$

Let $a=kx/y,b=ky/z,c=kz/x$, where all variables are positive reals. The inequality then becomes $$\sum_{\mathrm{cyc}} \frac{y}{kx+y+\frac{z}{k}}\geq \frac{3}{k+1+\frac{1}{k}}.$$By Cauchy-Schwarz, we have $$\left(\sum_{\mathrm{cyc}} \frac{y}{kx+y+\frac{z}{k}}\right)\left(\sum_{\mathrm{cyc}} y\left(kx+y+\frac{z}{k}\right)\right)\geq (x+y+z)^2,$$hence it suffices to prove that $$\left(k+\frac{1}{k}+1\right)(x+y+z)^2\geq 3\left(x^2+y^2+z^2+\left(k+\frac{1}{k}\right)(xy+yz+zx)\right).$$This rearranges to $$\left(k+\frac{1}{k}\right)(x^2+y^2+z^2-xy-yz-zx)\geq 2(x^2+y^2+z^2-xy-yz-zx),$$which is certainly true as $x^2+y^2+z^2\geq xy+yz+zx$ and $k+\tfrac{1}{k}\geq 2$ by AM-GM, so we're done. $\blacksquare$ Remark: This substitution is the lesser-known older brother of the $a=x/y,b=y/z,z=z/x$ substitution used when $abc=1$. Potentially helpful if you have a lot of terms which are degree $1$ and $-1$ that you can't seem to get rid of.

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This gives TSTST 2023/5 vibes. Let $x=a+\tfrac{1}{b}+1$, $y=b+\tfrac{1}{c}+1$, $z=c+\tfrac{1}{a}+1$, $n=\sqrt[3]{abc}+\tfrac{1}{\sqrt[3]{abc}}+1$, and $s=\tfrac{1}{x}+\tfrac{1}{y}+\tfrac{1}{z}$. It is equivalent to prove that $s \ge \tfrac{3}{n}$. Notice that \[abc+\frac{1}{abc}=(x-1)(y-1)(z-1)-(x-1)-(y-1)-(z-1)=(n-1)^3-3(n-1),\]which gives $xyz(1-s)=n^3-3n^2$. Since $n \ge 3$ by AM-GM, we have $xyz(1-s) \ge 0$, so $s \le 1$. Since $xyz \ge \tfrac{27}{s^3}$ by GM-HM, we have $\tfrac{27}{s^3}-\tfrac{27}{s^2} \le n^3-3n^2$. For $f(x)=x^3-3x^2=x^2(x-3)$, the above inequality is equivalent to $f(\tfrac{3}{s}) \le f(n)$. But notice that $\tfrac{3}{s},n \ge 3$ and $f(x)$ is increasing for $x \ge 3$, so $\tfrac{3}{s} \le n \implies s \ge \tfrac{3}{n}$, as desired.

Make the substitution $(a,b,c) = \left(\frac{kx}{y}, \frac{ky}{z}, \frac{kz}{x}\right)$. It then suffices to show \[\text{LHS} = \sum_{\text{cyc}} \frac{y^2}{kxy+y^2+\frac{yz}{k}} \ge \frac{(x+y+z)^2}{\left(k+\frac 1k\right) (xy+yz+zx) + (x^2+y^2+z^2)}\] is greater than $\frac{3}{k+\frac 1k+1}$. Expanding, our condition is equivalent to \[\left(k+\frac 1k-2\right)(x^2+y^2+z^2-xy-yz-zx) \ge 0,\] which is true. $\blacksquare$

Multiply both sides by $$\left(a+\frac 1b+1\right)\left(b+\frac 1c+1\right)\left(c+\frac 1a+1\right)\left(\sqrt[3]{abc}+\frac{1}{\sqrt[3]{abc}}+1\right)$$then $\text{LHS}-\text{RHS}$ is $$\frac{((abc)^{1/3}-1)^2}{(abc)^{4/3}} \left (\sum_{cyc} a^2b^2c-3(abc)^{5/3}+2\sum_{cyc} a^2bc-6(abc)^{4/3}+\sum_{cyc} a^2b-3abc+2\sum_{cyc} ab-6(abc)^{2/3}+\sum_{cyc} a-3(abc)^{1/3}\right)$$and the rest is obvious by AM-GM.