By using Cauchy inequality and Vasc ineqaulity, we have \[ LHS=\frac{x^4}{xz^3+x^3y}+\frac{y^4}{yx^3+y^3z}+\frac{z^4}{zy^3+z^3x} \] \[ \ge \frac{(x^2+y^2+z^2)^2}{2\left(x^3y+y^3z+z^3x\right)} \ge RHS \]

Bypassing Vasc's inequality, there is a simple proof?

Of course, I hope that my second solution is simple enough. By using AM-GM inequality, we have \[ LHS \ge \sum\limits_{cyc}{\frac{x^3}{\frac{2x^3+y^3}{3}+z^3}}=\sum\limits_{cyc}{\frac{3x^3}{2x^3+y^3+3z^3}} \] Now, let $ a=x^3,b=y^3,c=z^3 $, then it suffice to show \[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge \frac{1}{2} \] By using Cauchy inequality, \[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge {\frac{(a+b+c)^2}{\sum\limits_{cyc}{\left(2a^2+ab+3ac\right)}}}=\frac{1}{2} \]

Yeah. Very nice. Thank BSJL.

By Cauchy-Schwarz, we have \[\left(\frac{x^3}{z^3 + x^2y} + \frac{y^3}{x^3 + y^2z} + \frac{z^3}{y^3 + z^2x}\right) \ge \frac{\left(x^3 + y^3 + z^3\right)^2}{x^3\left(z^3 + x^2y\right) + y^3\left(x^3 + y^2z\right) + z^3\left(y^3 + z^2x\right)} .\] It remains to prove that \[2\left(x^3 + y^3 + z^3\right)^2 \ge 3\bigg(x^3\left(z^3 + x^2y\right) + y^3\left(x^3 + y^2z\right) + z^3\left(y^3 + z^2x\right)\bigg),\] which is equivalent to \[2\left(x^6 + y^6 + z^6\right) + \left(x^3y^3 + y^3z^3 + z^3x^3\right) \ge 3\left(x^5y + y^5z + z^5x\right).\] This follows from AM-GM: \[\text{LHS} = \left(x^6 + x^6 + x^3y^3\right) + \left(y^6 + y^6 + y^3z^3\right) + \left(z^6 + z^6 + z^3x^3\right) \ge 3x^5y + 3y^5z + 3z^5x. \; \square\]

Yes it is my solution BSJL wrote: Of course, I hope that my second solution is simple enough. By using AM-GM inequality, we have \[ LHS \ge \sum\limits_{cyc}{\frac{x^3}{\frac{2x^3+y^3}{3}+z^3}}=\sum\limits_{cyc}{\frac{3x^3}{2x^3+y^3+3z^3}} \] Now, let $ a=x^3,b=y^3,c=z^3 $, then it suffice to show \[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge \frac{1}{2} \] By using Cauchy inequality, \[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge {\frac{(a+b+c)^2}{\sum\limits_{cyc}{\left(2a^2+ab+3ac\right)}}}=\frac{1}{2} \]

ComplexPhi wrote: Let $x$,$y$,$z>0$ . Show that : $$\frac{x^3}{z^3+x^2y}+\frac{y^3}{x^3+y^2z}+\frac{z^3}{y^3+z^2x} \geq \frac{3}{2}$$ Let $x$ , $y$ , $z>0$ . Show that : $$\frac{x^4}{z^4+x^3y}+\frac{y^4}{x^4+y^3z}+\frac{z^4}{y^4+z^3x} \geq \frac{3}{2}$$

BSJL wrote:

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\[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge \frac{1}{2} \]

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Let $a ,b,c,d>0$ . Show that : $$\frac{a}{4a+5b+c+5d}+ \frac{b}{4b+5c+d+5a}+\frac{c}{4c+5d+a+5b}+\frac{d}{4d+5a+b+5c}\leq \frac{4}{5}$$

sqing wrote: Let $a ,b,c,d>0$ . Show that : $$\frac{a}{4a+5b+c+5d}+ \frac{b}{4b+5c+d+5a}+\frac{c}{4c+5d+a+5b}+\frac{d}{4d+5a+b+5c}\leq \frac{4}{5}$$ $$\sum_{cyc}\frac{a}{4a+5b+c+5d}\leq\frac{2}{5}$$a bit of stronger.

Thanks.

BSJL wrote: \[ \sum\limits_{cyc}{\frac{a}{2a+b+3c}} \ge \frac{1}{2} \] Prove that \[ \frac{a}{2a+b+3c} +\frac{b}{2b+c+3a}+\frac{c}{2c+a+3b}\geq \frac{1}{2} \]for all positive real numbers $a,b,c.$ Prove that \[ \frac{a}{b+2c+3d} +\frac{b}{c+2d+3a} +\frac{c}{d+2a+3b}+ \frac{d}{a+2b+3c} \geq \frac{2}{3} \]for all positive real numbers $a,b,c,d.$

sqing's second one: $\sum_{\text{cyc}}\frac a{b+2c+3d}=\sum_{\text{cyc}}\frac{a^2}{ab+2ac+3ad}\overset{\text{T2}}\ge\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}$ and it suffices to show $\frac{(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}\ge\frac23$. This is true iff $3(a+b+c+d)^2\ge8\sum_{\text{sym}}ab\Leftrightarrow4\left(a^2+b^2+c^2+d^2\right)\ge(a+b+c+d)^2$, which is obvious by C-S.

Easy inequality $\sum \frac{x^3}{z^3+x^2y}=\sum \frac{x^4}{z^3x+x^3y}\ge ^{T2}\frac{(x^2+y^2+z^2)^2}{2(\sum x^3y)}\ge \frac{3}{2}$ If ans only if $x^2+y^2+z^2\ge \sqrt{3(\sum x^3y}$ If and only if $ = \frac {1}{2}\left( \left( a^{2} - 2ab + bc - c^{2} + ca\right) ^2 + \left(b^{2} - 2bc + ca - a^{2} + ab\right)^2 + \left( c^{2} - 2ca + ab - b^{2} + bc\right)^2 \right)$