hello, can you fix your $\LaTeX$ please. Sonnhard.

Am I mistaken?

The equality $xy+yz+zx=3xyz$ is equivalent to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$. We have to prove that \[x^{2}y+y^{2}z+z^{2}x+3\ge 2x+2y+2z\] We will use above equality together with AM-GM inequality: \[x^{2}y+y^{2}z+z^{2}x+3=x^{2}y+y^{2}z+z^{2}x+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=(x^{2}y+\frac{1}{y})+(y^{2}z+\frac{1}{z})+(z^{2}x+\frac{1}{x})\ge 2x+2y+2z\]. Done!

Actually, much sharper, harder and more beautiful inequality holds: For $a,b,c>0$, such that $a+b+c=3$, prove: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge a^2+b^2+c^2 \quad (1)$ At first glance it may not be obvious how it is related to given inequality, but if we denote: $x=\frac{1}{a}, \; y=\frac{1}{b}, \; z=\frac{1}{c}$ in original inequality, condition becomes: $a+b+c=3$ and problem is equivalent to: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3abc \ge 2(ab+ac+bc)$ Now, from $(1)$ we have: $LHS\ge a^2+b^2+c^2+3abc$, so after homogenization, it remains to prove: $a^2+b^2+c^2+\frac{9abc}{a+b+c} \ge 2(ab+ac+bc)$ But that is just Schur's inequality of degree 3. $\blacksquare$

shivangjindal wrote: Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[ x^2y+y^2z+z^2x \ge 2(x+y+z)-3 \] and determine when equality holds. Refinement Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz.$ Prove that \[2x+2y+2z+3M\ge {{x}^{2}}y+{{y}^{2}}z+{{z}^{2}}x+3\ge 2x+2y+2z+3m,\]where \[M=\max \left\{ {{\left( xy-1 \right)}^{2}},{{\left( yz-1 \right)}^{2}},{{\left( zx-1 \right)}^{2}} \right\},\] \[m=\min \left\{ {{\left( xy-1 \right)}^{2}},{{\left( yz-1 \right)}^{2}},{{\left( zx-1 \right)}^{2}} \right\}.\]

Generalization 1 Let $n,x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[{{x}^{n+1}}{{y}^{n}}+{{y}^{n+1}}{{z}^{n}}+{{z}^{n+1}}{{x}^{n}}\ge \left( n+1 \right)\left( x+y+z \right)-3n\] and determine when equality holds.

By AM-GM $\sum_{cyc} x^{n+1}y^n + 3n = \sum_{cyc} x^{n+1}y^n + n\sum \dfrac {1}{x} =$ $ \sum_{cyc} \left (x^{n+1}y^n + n\dfrac {1} {y}\right ) \geq$ $ \sum (n+1)x =$ $ (n+1)\sum x$. Equality obviously holds if and only if $x=y=z=1$.

Generalization 2 Let $n,x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. If $k\in \left[ 0,n \right],$prove that \[{{x}^{n+1}}{{y}^{k}}+{{y}^{n+1}}{{z}^{k}}+{{z}^{n+1}}{{x}^{k}}\ge \left( n+1 \right)\left( x+y+z \right)-3n\] and determine when equality holds. Determine ${{k}_{\min }},{{k}_{\max }}$such that \[{{x}^{n+1}}{{y}^{k}}+{{y}^{n+1}}{{z}^{k}}+{{z}^{n+1}}{{x}^{k}}\ge \left( n+1 \right)\left( x+y+z \right)-3n,\forall x,y,z>0,xy+yz+xz=3xyz\]

We take $xy=a^2$, $yz=b^2$, $zx=c^2$ then $a^2+b^2+c^2=3abc$ $x=\frac{ac}{b}$, $y=\frac{ab}{c}$, $z=\frac{bc}{a}$ then it is equivalent to $\frac{a^4c^2+b^4a^2+c^4b^2}{abc}\geq2\frac{a^2c^2+b^2a^2+c^2b^2}{abc}-3$ $\frac{a^4c^2+b^4a^2+c^4b^2+a^2+b^2+c^2}{abc}\geq2\frac{a^2c^2+b^2a^2+c^2b^2}{abc}$ which is true by AM-GM

Condition $xy+yz+zx=3xyz$ is equivalent to $\sum{1\over x}=3$. Therefore from CS we get $\sum x^2y=\sum\frac{x^2}{\frac{1}{y}}\ge\frac{(\sum x)^2}{3}$. But ${(\sum x)^2\over3}\ge2\sum x -3$, because it's equivalent to $(3-\sum x)^2\ge0$. Q.E.D.

$x^2y+y^2z+z^2x=xyz\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)\geq xyz\cdot \frac{(x+y+z)^2}{xy+yz+zx}$ $=\frac{1}{3}(x+y+z)^2\geq 2(x+y+z)-3$, which is true by the tangent line at $(3,\ 3)$ on the parabola $y=\frac 13t^2.$

The following inequality is true. Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[ x^4y+y^4z+z^4x \ge \frac{2}{3}(x+y+z)^2-3. \]

The same method(s) work. By AM-GM, then Cauchy-Schwarz $\sum_{cyc} x^{4}y + 3 = \sum_{cyc} x^{4}y + \sum \dfrac {1}{x} =$ $ \sum_{cyc} \left (x^{4}y + \dfrac {1} {y}\right ) \geq$ $ \sum 2x^2 =$ $2\sum x^2 \geq \dfrac {2}{3} \left (\sum x\right )^2$. Equality holds if and only if $x=y=z=1$.

sqing wrote: The following inequality is true. Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[ x^4y+y^4z+z^4x \ge \frac{2}{3}(x+y+z)^2-3. \] Refinement Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[{{x}^{4}}y+{{y}^{4}}z+{{z}^{4}}x\ge \frac{{{\left( x+y+z \right)}^{4}}}{27}\ge \frac{2}{3}{{\left( x+y+z \right)}^{2}}-3\ge 4\left( x+y+z \right)-9\] and determine when equality holds.

Generalization 4 Let $n,x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. If $k\in \left[ 0,n \right],n\ge 1$,prove that \[{{x}^{n+1}}{{y}^{k}}+{{y}^{n+1}}{{z}^{k}}+{{z}^{n+1}}{{x}^{k}}\ge \frac{{{\left( x+y+z \right)}^{n+1}}}{{{3}^{n}}}\ge \frac{2}{{{3}^{\frac{n-1}{2}}}}\sqrt{{{\left( x+y+z \right)}^{n+1}}}-3\ge \left( n+1 \right)\left( x+y+z \right)-3n\] and determine when equality holds.

\[LET\: \: x=\frac{1}{a}\, \: \: y=\frac{1}{b}\: \: z=\frac{1}{c}\: \: THEN\: \: \: a+b+c=1 \: \: \: PROVE \: \: \: \frac{1}{a^2b}+\frac{1}{b^2c}+\frac{1}{c^2a}\geq 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3\: \: \: \: \: \: \: \: \: \: \sum \frac{1}{a^2b}+\sum a=\sum (\frac{1}{a^2b}+b)\geq 2(\sum \frac{1}{a})\]

Ummm... Solution seems to simple; might be wrong. 2-liner. $(\frac{x^2y+y^2z+z^2x}{3xyz})(3xyz)=(\frac{1}{3})(\frac{x}{z}+\frac{y}{x}+\frac{z}{y})(xy+xz+yz) \ge (\frac{1}{3})(x+y+z)^2$. Now it remains to prove that $\frac{1}{3}(x+y+z)^2 \ge 2(x+y+z)-3$, which transforms to $((x+y+z)-3)^2 \ge 0$, which is true.

\begin{align*} xy + yz + zx &= 3xyz \\ \frac {1} {x} + \frac {1} {y} + \frac {1} {z} &= 3 \\ ( ( x + y + z ) - 3)^2 &\geq 0 \\ ( x + y + z)^2 - 6 (x + y + z) + 9 &\geq 0 \\ ( x + y + z)^2 &\geq 3\cdot ( 2(x + y + z) - 3 ) \\ \left ( \frac {1} {x} + \frac {1} {y} + \frac {1} {z} \right )( x^2y + y^2z + z^2x) &\geq (x + y + z)^2 \geq 3\cdot ( 2(x + y + z) - 3 ) \\ \left ( \frac {1} {x} + \frac {1} {y} + \frac {1} {z} \right )( x^2y + y^2z + z^2x) &\geq 3\cdot ( 2(x + y + z) - 3 )\\ x^2y + y^2z + z^2x &\geq 2(x + y + z) - 3 \end{align*}

The first xy + yz + zx = 3xyz implies that 1/x + 1/y + 1/z = 3. Move the 3 from the RHS to the LHS, and substitute the above form, we get that x^2 *y + y^2 *z + z^2 *x + 1/x + 1/y + 1/z ≥ 2(x+y+z). You get x^2 *y + 1/y ≥ 2x, and so on for the y^2*z + 1/z ≥ 2y and z^2 * x + 1/x ≥ 2z. Adding, you get the expression you want to prove.

shivangjindal wrote: Let $x,y$ and $z$ be positive real numbers such that $xy+yz+xz=3xyz$. Prove that \[ x^2y+y^2z+z^2x \ge 2(x+y+z)-3 \]and determine when equality holds. UK - David Monk Generalization 5 For $x,y,z,k\in \mathbf{R^+}$ such that $xy+yz+zx=kxyz$. Prove the following $$x^{n+1}y^n+y^{n+1}z^n+z^{n+1}x^n\geq (n+1)(x+y+z)-kn$$ Determine equality case.

Generalization 6 For $a_{1},a_{2},\cdots,a_{p},k\in \mathbf{R^+}$ such that $\sum_{i=1-> p}{\frac{a_{1}a_{2}\cdots a_{n}}{a_{i}}}=k(a_{1}a_{2}\cdots a_{p})$ Prove the following $$a_{1}^{n+1}a_{2}^{n}+a_{2}^{n+1}a_{3}^{n}+\cdots+a_{p-1}^{n+1}a_{p}^{n}+a_{p}^{n+1}a_{1}^{n}\geq (n+1)(a_{1}+a_{2}+\cdots+a_{p})-kn$$ Determine the equality case.

Generalized BMO 2014 #1

Generalization 6 -Solution $$\sum_{i=1\to p}{\frac{a_{1}a_{2}\cdots a_{n}}{a_{i}}}=k(a_{1}a_{2}\cdots a_{p})\Rightarrow \sum_{cyc}{\dfrac{1}{a_{1}}}=k$$By homogenising $k$ with the given expression $$\sum_{cyc}{a_{1}^{n+1}a_{2}^{n}}\geq (n+1)(a_{1}+a_{2}+\cdots+a_{p})-kn =\sum_{cyc}{a_{1}^{n+1}a_{2}^{n}}+n\left(\sum_{cyc}{\dfrac{1}{a_{1}}}\right)$$$$=\sum_{cyc}{\left(a_{1}^{n+1}a_{2}^{n}+\overbrace{\dfrac{1}{a_{2}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{2}}}^{n}\right)}\overbrace{\geq}^{AM-GM} \sum_{cyc}{(n+1)a_{1}}=(n+1)(a_{1}+a_{2}+\cdots+a_{n})$$

By Cauchy-Schwarz, $$(x^2y+y^2z+z^2x) \left(\frac{1}{y}+\frac{1}{z}+\frac{1}{x}\right) \ge (x+y+z) ^2$$$$\Rightarrow x^2y+y^2z+z^2x \ge \frac{(x+y+z) ^2}{3} $$Let $x+y+z=t$, it suffices to prove $\frac{t^2}{3}\ge 2t-3$ which is equivalent to $\frac{(t-3)^2}{3}\ge 0 $.

Let $ x,y $ and $ z $ be positive real numbers such that $ x^2y+y^2z+z^2x+3= 2(x+y+z) $. Prove that$$xy+yz+xz\geq 3xyz$$

first idea immediately worked

Let $a = \frac{1}{x}$, $b = \frac{1}{y}$, and $c = \frac{1}{z}$. The given can then be written as \[a + b + c = 3\] and the inequality can be written as \[\sum_{\text{cyc}} \frac{1}{a^2b} \geq 2\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) - 3\]\[\frac{1}{(abc)^2} \sum_{\text{cyc}} ab^2 \geq \frac{2\left(ab + bc + ca\right)}{abc} - 3\]\[\sum_{\text{cyc}} ab^2 \geq 2abc(ab + bc + ca) - 3(abc)^2\] Using the given, the inequality becomes \[\sum_{\text{cyc}} ab^2 + (a+b+c)(abc)^2 \geq 2abc(ab + bc + ca)\] From AM-GM, we know that $ab^2 + a^3b^2c^2 \geq 2a^2b^2c$. Thus, \[\sum_{\text{cyc}} 2a^2b^2c \geq 2abc(ab + bc + ca)\] as desired.

cappucher wrote: Thus, \[\sum_{\text{cyc}} 2a^2b^2c \geq 2abc(a + b + c)\] as desired. I think $ab+ac+bc\geq a+b+c$ is not true

Here is an extension: General Case Suppose that positive reals $(a)_1^n$ satisfy $\sum\limits_{cyc}{a_1a_2\cdots a_{n-1}}=k\prod{a_1}$. Prove $$\sum_{cyc}{a_1^{p+1}a_2^p}\geq \left(p+1\right)\sum_{cyc}{a_1}-pk$$ holds for $k\in R^{+}$.

we see that $\frac{x^2y+y^2z+z^2x}{xyz}=3 \Rightarrow \sum_{\text{cyc}} \frac{1}{x}$ now in our inequality let's switch side of 3 $x^2y+y^2z+z^2x+3 \geq 2(x+y+z)$ now net's put our term as $x^2y+y^2z+z^2x+\sum_{\text{cyc}} \frac{1}{x} \geq 2(x+y+z)$ we can write it as $\Rightarrow \sum_{\text{cyc}}\frac{x^2y^2}{y}+\sum_{\text{cyc}} \frac{1}{x} \geq 2(x+y+z) $ $\Rightarrow \sum_{\text{cyc}}\frac{x^2y^2+1}{y} \geq 2(x+y+z)$ $RHS=\frac{2xy}{y}+\frac{2yz}{z}+\frac{2zx}{x}$ now we know that $(a-b)^2 \geq 0 $ similarly in our ineq $x^2y^2+1 \geq 2xy$ and we are done analogously other's also true and equeality holds when $x=y=z \Rightarrow x=y=z=1$ $\blacksquare$

note $3 = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ so multiple the LHS by this value and Cauchy-schwarz: $(\frac{1}{y}+\frac{1}{z}+\frac{1}{x})(x^2y+y^2z+z^2x) \ge (x+y+z)^2$, and we now wish to show $(x+y+z)^2 \ge 6(x+y+z)-9$. This is simply $(x+y+z-3)^2 \ge 0$

By the conditions ,we have$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$$Hence ,it suffices to show,$$\sum_{cyc}{x^2y}\ge\sum_{cyc}{(2x-\frac{1}{y})}$$which is equivalent to $$\sum_{cyc}{x^2y^2}\ge\sum_{cyc}{(2xy-1)}$$and it’s clear by AM-GM inequality. The equality holds when $x=y=z=1$. QED!