Simple bashing gives the expression like $2xyz(x^3+y^3+z^3)+(x^2y^4+y^2z^4+z^2x^4)\geq xyz(xy(x+y)+yz(y+z)+zx(z+x)) +3x^2y^2z^2$ well $2xyz(x^3+y^3+z^3)\geq xyz(xy(x+y)+yz(y+z)+zx(z+x)$ and $(x^2y^4+y^2z^4+z^2x^4)\geq 3x^2y^2z^2$.. so done!!

emregirgin35 wrote: For all positive real numbers $x, y, z$, show that $ \frac{x(2x-y)}{y(2z+x)}+\frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)} \geq 1$ is true. after expanding is equivalent to $8\sum{x^4yz}+4\sum_{cyc} {x^2y^4} \ge 12x^2y^2z^2 +4\sum{x^3y^2z}$ which follows by $8\sum{x^4yz} \ge 4\sum{x^3y^2z}$ (Muirhead) $4\sum_{cyc} {x^2y^4} \ge 12x^2y^2z^2 $ (AMGM)

we : $ \sum \frac{x(2x-y)}{y(2z+x)}\geq 3\frac{2\sum (x^{2}-xy)}{2\sum xy+\sum xy}=3\frac{2\sum x^{2}-\sum xy}{3\sum xy}\geq 1\Leftrightarrow 2\sum x^{2}\geq 2\sum xy $

nicusorz wrote: we : $ \sum \frac{x(2x-y)}{y(2z+x)}\geq 3\frac{2\sum (x^{2}-xy)}{2\sum xy+\sum xy}=3\frac{2\sum x^{2}-\sum xy}{3\sum xy}\geq 1\Leftrightarrow 2\sum x^{2}\geq 2\sum xy $ Hi, Where does the first inequality come? Also \[\sum \frac{a_i}{b_i}\ge n\frac{\sum a_i}{\sum b_i}\] isn't always true for subcase $n=3$.

emregirgin35 wrote: For all positive real numbers $x, y, z$, show that $ \frac{x(2x-y)}{y(2z+x)}+\frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)} \geq 1$ is true. A solution without expanding: \[\sum\left(\frac{x(2x-y)}{y(2z+x)}+1\right)=\sum \frac{2x^2+2yz}{y(2z+x)}\] So it is enough to show \[\sum\frac{x^2}{y(2z+x)}+\sum\frac{z}{2z+x} \ge 2\] But note that By twice C-S \begin{align*}\sum\frac{x^2}{y(2z+x)}+\sum\frac{z^2}{2z^2+zx} & \ge \frac{(x+y+z)^2}{3(xy+yz+zx)}+\frac{(x+y+z)^2}{2\sum x^2 +\sum xy} \\ & \ge \frac{4(x+y+z)^2}{3\sum xy+2\sum x^2 +\sum xy}=2\end{align*}

It is equivalent to For all positive real numbers $x, y, z$ such that $xyz=1 $ , show that \[ \frac{x^3+1}{x^2y+2}+\frac{y^3+1}{y^2z+2}+\frac{z^3+1}{z^2x+2}\geq 2\] is true .

It's equivalent to \[ \sum_{\text{cyc}} \frac{x^2+yz}{2yz+xy} \ge 2. \] Now by Cauchy in Engel form (called Titu in the USA), we obtain \begin{align*} \sum_{\text{cyc}} \frac{x^2+yz}{2yz+xy} &= \sum_{\text{cyc}} \frac{x^4+x^2yz}{2x^2yz+x^3y} \\ &\ge \frac{(x^2+y^2+z^2)^2 + \left( x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy} \right)^2}{x^3y+y^3z+z^3x+2xyz(x+y+z)} \\ &\ge \frac{3(x^3y+y^3z+z^3x) + xyz(x+y+z) + 2\sum_{\text{cyc}} x^{\frac{3}{2}}y^{\frac{3}{2}}z}{x^3y+y^3z+z^3x+2xyz(x+y+z)} \\ &= \frac{2(x^3y+y^3z+z^3x)+4xyz(x+y+z) + \sum_{\text{cyc}} \left( x^3y + 2x^{\frac{3}{2}}yz^{\frac{3}{2}} - 3x^2yz \right)}{x^3y+y^3z+z^3x+2xyz(x+y+z)} \\ &\ge \frac{2(x^3y+y^3z+z^3x)+4xyz(x+y+z)}{x^3y+y^3z+z^3x+2xyz(x+y+z)} \\ &= 2. \end{align*} Here, we have applied Vasc's Inequality followed by AM-GM as $x^3y + 2x^{\frac{3}{2}}yz^{\frac{3}{2}} \ge 3x^2yz$.

v_Enhance wrote: It's equivalent to \[ \sum_{\text{cyc}} \frac{x^2+yz}{2yz+xy} \ge 2. \] I think there is nice solutions for \[ \sum_{\text{cyc}} \frac{x^2+6yz}{xy+2yz} \ge 7 \] and \[ \sum_{\text{cyc}} \frac{x^2+7yz}{xy+yz} \ge 12. \]

Vasc wrote: v_Enhance wrote: It's equivalent to \[ \sum_{\text{cyc}} \frac{x^2+yz}{2yz+xy} \ge 2. \] I think there is nice solutions for \[ M_1=\sum_{\text{cyc}} \frac{x^2+6yz}{xy+2yz}-7\ge 0 \] and \[ M_2=\sum_{\text{cyc}} \frac{x^2+7yz}{xy+yz}-12\ge 0. \] My bad prove is: \[M_1=\frac{P}{y(x+2z)z(2x+y)x(2y+z)}.\] $P=2x^2y^2(y-z)^2+3y^2(x-y)^2xz+\frac{3}{2}x^2(x-z)^2yz+4z^2(x-z)^2xy+\frac{1}{2}x^2(y-z)^2yz+2z^2(x^2+yz-2xy)^2+y^2(y-z)^2xz+\frac{5}{2}x^2(x-y)^2yz.$ and for $ M_2$ , $ P=\frac{1}{12}(xy^2+x^2z-2yz^2)^2+\frac{5}{3}z^2(x-y)^2xy+\frac{5}{3}x^2(y-z)^2yz+\frac{2}{3}y^2(x-z)^2xz+\frac{1}{3}y^2(z^2-2xz+xy)^2+x^2(x-z)^2yz+y^2(x-y)^2xz+z^2(y-z)^2xy+\frac{1}{3}z^2(x^2+yz-2xy)^2+\frac{7}{12}x^2(y^2-2yz+xz)^2.$ BQ

Vasc wrote: v_Enhance wrote: It's equivalent to \[ \sum_{\text{cyc}} \frac{x^2+yz}{2yz+xy} \ge 2. \] I think there is nice solutions for \[ \sum_{\text{cyc}} \frac{x^2+6yz}{xy+2yz} \ge 7 \] and \[ \sum_{\text{cyc}} \frac{x^2+7yz}{xy+yz} \ge 12. \] let$\frac{x^2+6yz} {xy+2yz}=a$ then $x^2+y^2+z^2-xy-yz-xz=(a+2c-7)xy+(2a+b-7)yz+ +(c+2b-7)zy$ so RHS is nonnegative for all $x,y,z$ so $a+2c-7\ge0 , 2a+b-7\ge0, c+2b-7\ge0$ by adding these $a+b+c\ge7$ similarly for the second one i think this method could be used to bound any fractional sum whose enumerators and denominators are linear combinations of $x^2,y^2,z^2,xy,yz,zy$ for all nonnegative reals $x,y,z$ with the adding that sometimes the upper bonds are obtained

abitofmath wrote: let$\frac{x^2+6yz} {xy+2yz}=a$ then $x^2+y^2+z^2-xy-yz-xz=(a+2c-7)xy+(2a+b-7)yz+ +(c+2b-7)zy$ so RHS is nonnegative for all $x,y,z$ so $a+2c-7\ge0 , 2a+b-7\ge0, c+2b-7\ge0$ by adding these $a+b+c\ge7$ You mean $a+2c\ge 7$ always holds? That is \[\frac{x^2+6yz}{xy+2yz}+2\frac{z^2+6xy}{zx+2xy} \ge 7\] which is false for $x=5, y=10, z=1$ I think in the equation you put $z=0$ and found $a+2c\ge 7$. But you forgot that $a,c$ are dependent on $z$.

you are right sayan it is a mestake actually $z=0$ would lead to $(2c+a)xy=x^2+y^2-xy$ and it comes out as nonsense thanks

The following inequality is also true. For all positive real numbers $x, y, z$, show that \[ \frac{x^2(2x-y)}{y^2(2z+x)}+\frac{y^2(2y-z)}{z^2(2x+y)}+\frac{z^2(2z-x)}{x^2(2y+z)} \geq 1.\]

sqing wrote: The following inequality is also true. For all positive real numbers $x, y, z$, show that \[ \frac{x^2(2x-y)}{y^2(2z+x)}+\frac{y^2(2y-z)}{z^2(2x+y)}+\frac{z^2(2z-x)}{x^2(2y+z)} \geq 1.\] It's equivalent to $\sum_{cyc}(4x^6z^3+8x^6z^2y-6x^4y^2z^2-3x^4z^3y^2-3x^3y^3z^3)\geq0$, which is true by AM-GM.

Let x,y,z be positive real numbers,k(>=1) is integer. Then, The following inequality is also true \[\sum\frac{x^k(2x-y)}{y^k(x+2z)}\ge 1\]

The following inequality is also true. For all positive real numbers $x, y, z$, show that\[\frac{x(3x-2y)}{y(2z+x)}+\frac{y(3y-2z)}{z(2x+y)}+\frac{z(3z-2x)}{x(2y+z)} \geq 1.\]

sqing wrote: The following inequality is also true. For all positive real numbers $x, y, z$, show that\[\frac{x(3x-2y)}{y(2z+x)}+\frac{y(3y-2z)}{z(2x+y)}+\frac{z(3z-2x)}{x(2y+z)} \geq 1.\] hello, with $y=x+u,z=x+u+v$ we get $6 u^6+u^5 (24 v+47 x)+u^4 \left(36 v^2+147 v x+123 x^2\right)+u^3 \left(24 v^3+189 v^2 x+293 v x^2+129 x^3\right)+u^2 \left(6 v^4+113 v^3 x+318 v^2 x^2+217 v x^3+47 x^4\right)+u \left(24 v^4 x+148 v^3 x^2+206 v^2 x^3+47 v x^4\right)+18 v^4 x^2+59 v^3 x^3+47 v^2 x^4\geq 0$ Sonnhard.

hello, or this here $6 u^6+24 u^5 v+36 u^4 v^2+24 u^3 v^3+6 u^2 v^4+x^4 \left(47 u^2+47 u v+47 v^2\right)+x^3 \left(129 u^3+217 u^2 v+206 u v^2+59 v^3\right)+x^2 \left(123 u^4+293 u^3 v+318 u^2 v^2+148 u v^3+18 v^4\right)+x \left(47 u^5+147 u^4 v+189 u^3 v^2+113 u^2 v^3+24 u v^4\right)\geq 0$ Sonnhard.

hello, or this here $6 u^6+24 u^5 v+36 u^4 v^2+24 u^3 v^3+6 u^2 v^4+x^4 \left(47 u^2+47 u v+47 v^2\right)+x^3 \left(129 u^3+217 u^2 v+206 u v^2+59 v^3\right)+x^2 \left(123 u^4+293 u^3 v+318 u^2 v^2+148 u v^3+18 v^4\right)+x \left(47 u^5+147 u^4 v+189 u^3 v^2+113 u^2 v^3+24 u v^4\right)\geq 0$ Sonnhard.

Here's a generalization http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=598893

emregirgin35 wrote: For all positive real numbers $x, y, z$, show that $ \frac{x(2x-y)}{y(2z+x)}+\frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)} \geq 1$ is true. \[Q=\frac{x(2x-y)}{y(2z+x)}+\frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)},\] \[\sum{\frac{x}{y}}\geq \frac{1}{2}Q+\frac{5}{2}\geq \sum{\frac{x+y}{y+z}}\] BQ

emregirgin35 wrote: For all positive real numbers $x, y, z$, show that $ \frac{x(2x-y)}{y(2z+x)}+\frac{y(2y-z)}{z(2x+y)}+\frac{z(2z-x)}{x(2y+z)} \geq 1$ is true. $$\sum \frac{2x^2}{y(2z+x)} \geq 1+ \sum \frac{xy}{2yz + xy} = 1+ \sum 1- \frac{2yz}{2yz+xy} = 4 -\sum \frac{2yz}{2yz+xy}$$$$\sum \frac{x^2+yz}{2yz+xy} \geq 2$$$$x^2+y^2+z^2 =A,xy+yz+zx=B$$$$(\sum \frac{x^2}{2yz+xy})(\sum 2yz+xy) \geq (x+y+z)^2$$$$\sum \frac{yz}{2yz+xy} = \sum \frac{z}{2z+x}$$$$(\sum \frac{z}{2z+x})(\sum 2z^2 + xz) \geq A^2$$$$\frac{A+2B}{3B} + \frac{A+2B}{2A+B} \geq 2$$$$(A-B)^2 \geq 0$$$\blacksquare$

For all positive real numbers $x, y, z$, show that$$\frac{x(3x-y)}{y(3z+x)}+\frac{y(3y-z)}{z(3x+y)}+\frac{z(3z-x)}{x(3y+z)}\geq \frac{3}{2}$$

(Hüseyin Emekçi) $\dfrac{x(2x-y)}{y(2z+x)}+\dfrac{y(2y-z)}{z(2x+y)}+\dfrac{z(2z-x)}{x(2y+z)}=\sum{\frac{2x^2}{2yz+xy}}-\sum{\frac{xy}{2yz+xy}}$ $\geq \frac{2(x+y+z)^2}{3(xy+yz+zx)}-\sum{\frac{xy}{2yz+xy}}$ $=\frac{2(x+y+z)^2}{3(xy+yz+zx)}-(3-\sum{\frac{2yz}{2yz+xy}}=\frac{2(x+y+z)^2}{3(xy+yz+zx)}-3+\sum{\frac{2yz}{2yz+xy}}\geq 1$ $=> \frac{(x+y+z)^2}{3(xy+yz+zx)}+\sum{\frac{yz}{2yz+xy}}=\frac{(x+y+z)^2}{3(xy+yz+zx)}+\frac{z}{2z+x}+\frac{x}{2x+y}+\frac{y}{2y+z}\geq 2$ $\frac{(x+y+z)^2}{3(xy+yz+zx)}+\frac{z}{2z+x}+\frac{x}{2x+y}+\frac{y}{2y+z}\geq \frac{(x+y+z)^2}{3(xy+yz+zx)}+\frac{(x+y+z)^2}{2(x^2+y^2+z^2)+xy+yz+zx}$ $=(x+y+z)^2\left[\frac{1}{3(xy+yz+zx)}+\frac{1}{2(x^2+y^2+z^2)+xy+yz+zx}\right]\geq (x+y+z)^2\left(\frac{4}{4(xy+yz+zx)+2(x^2+y^2+z^2)}\right)\geq 2$ $\frac{2(x+y+z)^2}{4(xy+yz+zx)+2(x^2+y^2+z^2)}\geq 1$. This expression is trivial. Proof completed.

sqing wrote: For all positive real numbers $x, y, z$, show that$$\frac{x(3x-y)}{y(3z+x)}+\frac{y(3y-z)}{z(3x+y)}+\frac{z(3z-x)}{x(3y+z)}\geq \frac{3}{2}$$

sqing wrote: For all positive real numbers $x, y, z$, show that$$\frac{x(3x-y)}{y(3z+x)}+\frac{y(3y-z)}{z(3x+y)}+\frac{z(3z-x)}{x(3y+z)}\geq \frac{3}{2}$$ I sent the solution Sqing, you can look at it.

(Hüseyin Emekçi) Generalised Form of This Type of Question: $x,y,z,a,b\in \mathbf{R^+}$. Then prove that $\frac{x(ax-by)}{y(az+bx)}+\frac{y(ay-bz)}{z(ax+by)}+\frac{z(az-bx)}{x(ay+bz)}\geq 3\frac{a-b}{b+c}$

The inequality is homogenous so we can assume that $xyz=1$. \[\sum{\frac{2x^2-xy}{xy+2yz}}\geq 1\overset{\text{by adding 1 to each term}}{\iff} \sum{\frac{x^2+yz}{xy+2yz}}\geq 2\]\[\sum{\frac{x^2+yz}{xy+2yz}}\geq 3\sqrt[3]{\frac{\Pi{(x^2+yz)}}{\Pi{(xy+2yz)}}}\overset{\text{xyz=1}}{=}3\sqrt[3]{\frac{\Pi{(x^3+1)}}{\Pi{(x+2z)}}}\overset{?}{\geq} 2\]\[27\Pi{(x^3+1)}\overset{?}{\geq}8\Pi{(x+2z)}\]\[54+27\sum{x^3y^3}+27\sum{x^3}\overset{?}{\geq}72+16\sum{xy^2}+32\sum{x^2y}\]\[27\sum{x^3y^3}+27\sum{x^3}\overset{?}{\geq}18+16\sum{xy^2}+32\sum{x^2y}\]We have \[\frac{33}{2}\sum{x^3}\geq 16\sum{xy^2}+\frac{1}{2}\sum{x^2y}\]by rearrangement inequality on $(x\geq y\geq z),(x^2\geq y^2\geq z^2)$. \[\frac{21}{2}(2\sum{x^3y^3}+\sum{x^3})\geq \frac{21}{2}(\sum{x^3y^3}+\sum{x^3}+\sum{1})\geq \frac{63}{2}\sum{x^2y}\]\[6\sum{x^3y^3}\geq 18\]By adding these $3$ inequalities we get the desired result.$\blacksquare$

For all positive real numbers $x, y, z$, show that $$ \frac{x(4x-y)}{y(4z+x)}+\frac{y(4y-z)}{z(4x+y)}+\frac{z(4z-x)}{x(4y+z)} \geq \frac{9}{5}$$$$ \frac{x(5x-y)}{y(5z+x)}+\frac{y(5y-z)}{z(5x+y)}+\frac{z(5z-x)}{x(5y+z)}\geq 2$$