Just a simple application of chebyshev inequality. \[ \sum_{cyc} \frac{x^3}{y^2+z^2} \ge \frac{1}{3}\sum_{cyc} x^3 .\sum_{cyc}\frac{1}{y^2+z^2}\] \[ \ge \frac{1}{3} .\frac{1}{3} \sum_{cyc} x.\sum_{cyc} x^2 .\frac{9}{2\sum_{cyc}x^2}\] \[ = \frac{x+y+z}{2}\]

mathisfun7 wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] Observe that: $\frac{x^3}{y^2+z^2}=x-\frac{x(y^2+z^2-x^2)}{y^2+z^2}$. $\ge x-\frac{x(y^2+z^2-x^2)}{2yz}$ by AM-GM. So it remains to show that: $x+y+z-\frac{1}{2}\sum \frac{x(y^2+z^2-x^2)}{yz}\ge \frac{1}{2}$ $\Leftrightarrow xyz(x+y+z)\ge \sum x^2(y^2+z^2-x^2)$ Which is true by expanding and Schur. (Thanks to infiniteturtle for this part. )

By Chebyshev : \[LHS \ge \frac{1}{3}(x+y+z)(\[\dfrac{x^2}{y^2+z^2}+\dfrac{y^2}{z^2+x^2}+\dfrac{z^2}{x^2+y^2}) \ge \dfrac{x+y+z}{2}\] http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=569530 Similar to the old inequality(Bulgaria Стара Загора 1988): Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge \dfrac{x+y+z}{2}.\]

sqing wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge \dfrac{x+y+z}{2}.\] Trivial AM-GM: $\frac{x^2}{y+z}+\frac{y+z}{4}\ge x$ and sum the similar inequalities we get the inequality.

mathisfun7 wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \frac{3}{2}\sqrt[12]{\dfrac{x^{12}+y^{12}+z^{12}}{3}}\] a bit of stronger.

mathisfun7 wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] Are you joking? It is too hard problem for such international olympiad "Baltic way" . Even the following is overused: \[\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}+\dfrac{z^3}{z^2+x^2}\ge \dfrac{x+y+z}{2}.\]

sqing wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge \dfrac{x+y+z}{2}.\] It also can be done by using Tittu's lemma or rearrangement.

WLOG let $x\le y \le z$. By the rearrangement inequality, \[ \frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2}\ge \frac{z^3}{y^2+z^2}+\frac{x^3}{z^2+x^2}+\frac{y^3}{x^2+y^2} \] \[ \frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2}\ge \frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}+\frac{x^3}{x^2+y^2} \] Summing these two we get \[ \frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2}\ge \frac{1}{2}(\frac{y^3+z^3}{y^2+z^2}+\frac{z^3+x^3}{z^2+x^2}+\frac{x^3+y^3}{x^2+y^2}) (*) \] By Muirhead, \[x^3+y^3\ge x^2y+xy^2\] Adding $x^3+y^3$ to both sides and rearranging terms gives, \[\frac{x^3+y^3}{x^2+y^2} \ge \frac{x+y}{2}\] Applying this to $(*)$ we get the desired inequality, and we're done. $\blacksquare$

Prove that the following inequality holds for all positive real numbers $x,y,z$: $\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$$\ge \frac{1}{2}(\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}+\frac{x^2+y^2}{x+y})\ge \frac{x+y+z}{2}.$

The sequences $(x^2,y^2,z^2)$ and $\left(\frac{1}{y+z},\frac{1}{z+x},\frac{1}{x+y}\right)$ are sorted the same way.So we get $\sum\frac{x^2}{y+z}\ge \sum\frac{y^2}{y+z}$ and $\sum\frac{x^2}{y+z}\ge \sum\frac{z^2}{y+z}$ adding these two inequalities we get the middle one and the last one is easy AM-GM using the lemma:$\frac{y^2+z^2}{y+z}\ge \frac{y+z}{2}$ and summing similar inequalities we get the desired one.

Generalization 1 Prove that the following inequality holds for all positive real numbers $x,y,z$ and $\alpha ,\beta >0$ \[\frac{{{x}^{\alpha +\beta }}}{{{y}^{\alpha }}+{{z}^{\alpha }}}+\frac{{{y}^{\alpha +\beta }}}{{{z}^{\alpha }}+{{x}^{\alpha }}}+\frac{{{z}^{\alpha +\beta }}}{{{x}^{\alpha }}+{{y}^{\alpha }}}\] \[\ge \frac{1}{2}\left( \frac{{{y}^{\alpha +\beta }}+{{z}^{\alpha +\beta }}}{{{y}^{\alpha }}+{{z}^{\alpha }}}+\frac{{{z}^{\alpha +\beta }}+{{x}^{\alpha +\beta }}}{{{z}^{\alpha }}+{{x}^{\alpha }}}+\frac{{{x}^{\alpha +\beta }}+{{y}^{\alpha +\beta }}}{{{x}^{\alpha }}+{{y}^{\alpha }}} \right)\ge \frac{{{x}^{\beta }}+{{y}^{\beta }}+{{z}^{\beta }}}{2}.\]

Generalization 2 Let $x,y,z$ positive real numbers and $f,g,h\colon (0,\infty)\to (0,\infty)$ with $f,g,h$ ascending functions,$h(t)=\frac{f(t)}{g(t)}$ Prove that $\frac{f(z)}{g(x)+g(y)}+\frac{f(x)}{g(y)+g(z)}+\frac{f(y)}{g(z)+g(x)}\ge $ $\frac{1}{2}\left( \frac{f(x)+f(y)}{g(x)+g(y)}+\frac{f(y)+f(z)}{g(y)+g(z)}+\frac{f(z)+f(x)}{g(z)+g(x)} \right)\ge \frac{h\left( x \right)+h\left( y \right)+h\left( z \right)}{2}$ Generalization 3 Let $x,y,z$ positive real numbers and $f,g,h\colon (0,\infty)\to (0,\infty)$ ,$h(t)=\frac{f(t)}{g(t)}$ with $f,g,h$ descending functions Prove that $\frac{f(z)}{g(x)+g(y)}+\frac{f(x)}{g(y)+g(z)}+\frac{f(y)}{g(z)+g(x)}\ge $ $\frac{1}{2}\left( \frac{f(x)+f(y)}{g(x)+g(y)}+\frac{f(y)+f(z)}{g(y)+g(z)}+\frac{f(z)+f(x)}{g(z)+g(x)} \right)\ge \frac{h\left( x \right)+h\left( y \right)+h\left( z \right)}{2}$

Generalization 4(reverse) Let $x,y,z$ positive real numbers and $f,g,h\colon (0,\infty)\to (0,\infty)$ ,$h(t)=\frac{f(t)}{g(t)}$ with a)$f,h$ descending functions and $g$ ascending functions or b)$f,h$ ascending functions and $g$ descending functions Prove that $\frac{f(z)}{g(x)+g(y)}+\frac{f(x)}{g(y)+g(z)}+\frac{f(y)}{g(z)+g(x)}\le $ $\frac{1}{2}\left( \frac{f(x)+f(y)}{g(x)+g(y)}+\frac{f(y)+f(z)}{g(y)+g(z)}+\frac{f(z)+f(x)}{g(z)+g(x)} \right)\le \frac{h\left( x \right)+h\left( y \right)+h\left( z \right)}{2}$

The following inequality is also true. Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^2y}{y^2+z^2}+\dfrac{y^2z}{z^2+x^2}+\dfrac{z^2x}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] \[\dfrac{x^2\sqrt{yz}}{y^2+z^2}+\dfrac{y^2\sqrt{zx}}{z^2+x^2}+\dfrac{z^2\sqrt{xy}}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\]

sqing wrote: The following inequality is also true. Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^2y}{y^2+z^2}+\dfrac{y^2z}{z^2+x^2}+\dfrac{z^2x}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] \[\dfrac{x^2\sqrt{yz}}{y^2+z^2}+\dfrac{y^2\sqrt{zx}}{z^2+x^2}+\dfrac{z^2\sqrt{xy}}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] Are wrong . $z\searrow 0,x=1,y=2$

Solution. By Cauchy-Schwarz, \[\frac{a+b+c}{2}=\frac{(a+b+c)^2}{(a+b)+(b+c)+(c+a)}\le \sum \frac{a^2}{b+c}.\] It suffices to prove \[\sum \frac{a^3}{b^2+c^2}\ge \sum \frac{a^2}{b+c}\] \[\Leftrightarrow \sum \frac{a^2b(a-b)-a^2c(c-a)}{(b+c)(b^2+c^2)}\ge 0\] \[\Leftrightarrow \sum \frac{a^2b(a-b)}{(b+c)(b^2+c^2)}-\frac{ab^2(a-b)}{(c+a)(c^2+a^2)}\ge 0\] \[\Leftrightarrow \sum (a-b)\frac{a^2b(c+a)(c^2+a^2)-ab^2(b+c)(b^2+c^2)}{(b+c)(b^2+c^2)(c+a)(c^2+a^2)}\ge 0\] \[\Leftrightarrow \sum (a-b)^2\frac{ab(\sum a^3+\sum ab(a+b)+abc)}{(b+c)(b^2+c^2)(c+a)(c^2+a^2)}\ge 0\] which is obvious. $\blacksquare$ See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=566615 Arkan

$ x-\frac{x(y^2+z^2-x^2)}{2yz} $have not $ \frac{x^3}{y^2+z^2}\ge x-\frac{x(y^2+z^2-x^2)}{2yz} $ $ y^2+z^2-x^2 $＞0？

Inequality from Hungary Let $x,y,z$ be positive real numbers such that $xyz=1$ .:Prove that the following inequality \[\dfrac{y^3+z^3}{x^2+xy+y^2}+\dfrac{z^3+x^3}{y^2+yz+z^2}+\dfrac{x^3+y^3}{z^2+zx+x^2}\ge 2.\]

Good inequality WLOG $x\le y\le z$ by Chebishev we have $\sum\frac{x^3}{y^2+z^2}\ge \frac{(x^3+y^3+z^3)(\sum\frac{1}{y^2+z^2})}{3}$ here again Chebishev $x^3+y^3+z^3\ge \frac{(\sum x)(\sum x^2)}{3}$ after we ll use C-S $(\sum x^2+y^2)(\sum \frac{1}{x^2+y^2})\ge 9$ the rest is just calculation.

AlgebraFC wrote: Stronger: Show that for all positive reals $x, y, z$, \[\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq x+y+z.\] This is Stronger，Cauchy and schur inequality

AlgebraFC wrote: Stronger: Show that for all positive reals $x, y, z$, \[\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq x+y+z.\] By C-S $$\sum_{cyc}\frac{x^3}{y^2-yz+z^2}\geq \frac{(x^2+y^2+z^2)^2}{x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)-3xyz}$$Hence we only need to prove $$(x^2+y^2+z^2)^2\ge [x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)-3xyz](x+y+z)$$which is equivalent to $$\sum_{cyc}x^2(x-y)(x-z)\ge 0$$Done.

Wrong post.

weirdo37 wrote: AlgebraFC wrote: Stronger: Show that for all positive reals $x, y, z$, \[\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq x+y+z.\] Hello, your inequality is weaker since $y^2-xy+x^2\leq \frac{1}{2}(y^2+x^2)$ (applying this to the denominators and we've got the initial inequality) $y^2-xy+x^2\ge \frac{1}{2}(y^2+x^2)$

Murad.Aghazade wrote: weirdo37 wrote: AlgebraFC wrote: Stronger: Show that for all positive reals $x, y, z$, \[\frac{x^3}{y^2-yz+z^2}+\frac{y^3}{z^2-zx+x^2}+\frac{z^3}{x^2-xy+y^2}\geq x+y+z.\] Hello, your inequality is weaker since $y^2-xy+x^2\leq \frac{1}{2}(y^2+x^2)$ (applying this to the denominators and we've got the initial inequality) $y^2-xy+x^2\ge \frac{1}{2}(y^2+x^2)$ oops. sorry, deleting this post.

Clearing the denominator, we get $S(7,0,0)+S(5,2,0) \geq S(4,3,0)+S(4,2,1)$ Obviously this holds by Muirhead inequality

mathisfun7 wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \ge \frac{(x^2+y^2+z^2)^2}{x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)} $ $\ge \frac{(x^2+y^2+z^2)^2}{\frac{2}{3}(x^2+y^2+z^2)(x+y+z)}=\frac{3(x^2+y^2+z^2)}{2(x+y+z)}$ $\ge\frac{(x+y+z)^2}{2(x+y+z)}=\frac{x+y+z}{2}$

mathisfun7 wrote: Prove that the following inequality holds for all positive real numbers $x,y,z$: \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\] \[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2} \geq \frac{1}{3}(x+y+z)(\dfrac{x^2}{y^2+z^2}+\dfrac{y^2}{z^2+x^2}+\dfrac{z^2}{x^2+y^2}) \geq \frac{x+y+z}{2}\]

Let $f(t) = \sum_{cyc} \frac{x^{t+1}}{y^t+z^t}, \quad t \in [0, \infty)$ If my computations are correct then $f'(t) = \frac{\sum (\ln x - \ln y) x^t y^t (x^t+y^t)^2 [(x^{2t+1} - y^{2t+1}) + z^{2t}(x-y)+2z^t(x^{t+1}-y^{t+1})]}{(x^t+y^t)^2(y^t+z^t)^2(z^t+x^t)^2} \geq 0$ because $(\ln x - \ln y)$ and $[(x^{2t+1} - y^{2t+1}) + z^{2t}(x-y)+2z^t(x^{t+1}-y^{t+1})]$ have the same sign (the sign of $x-y$) Thus $f$ is increasing and this solve initial question ( and several others on this thread)

Clearing the denominator, we get $2(\sum_{cyc}x^7)+(\sum_{cyc}x^5y^2+x^5z^2)\geq (\sum_{cyc}x^4y^3+\sum_{cyc}x^4z^3)+(\sum_{cyc}x^4y^2z+\sum_{cyc}x^4yz^2)$ This holds by Muirhead inequality

Here's a solution with vasc's inequality(here , all sums are cyclic): WLOG $\sum x^3y \ge \sum xy^3$ .By T2 lemma: $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \ge \frac{(x^2+y^2+z^2)^2}{x(y^2+z^2)+y(z^2+x^2)+z(x^2+y^2)} $ which is equivalent to: $2(x^2 + y^2 + z^2)^2 \ge 2\sum x^2y^2 + \sum x^3y + \sum xy^3 + \sum xyz^2 +\sum xy^2z$ by vasc's inequality: $LHS \ge 6 \sum x^3y$ .now , we are done by rearrangement inequality and $\sum x^3y \ge \sum xy^3$ .

After expansion, we need \begin{align*} 2x^7+2x^3y^2z^2+2x^5z^2+2x^5y^2\\+2y^7+2x^2y^3z^2+2y^5z^2+2y^5x^2\\+2z^7+2x^2y^2z^3+2z^5y^2+2z^5x^2\geq\\ x^3z^4+x^5z^2+x^5y^2+2x^3y^2z^2+xy^2z^4+x^3y^4+xy^4z^2\\+x^2yz^4+x^4yz^2+x^4y^3+2x^2y^3z^2+y^3z^4+y^5x^2+y^5z^2\\+x^2z^5+x^4z^3+x^4y^2z+2x^2y^2z^3+y^2z^5+y^4x^2z+y^4z^3, \end{align*}which is actually equivalent to $$\sum_{sym} x^7y^0z^0+\sum_{sym}x^5y^2z^0\geq \sum_{sym} x^4y^3z^0+\sum_{sym} x^4y^2z^1,$$which is true by Muirhead.

sos is easy in doing this problem

I think also this problem is true! $$\sum \frac{a^4}{a^3+2b^3} \geq \frac{a+b+c+d}{3}$$