Parker Pitts

2022-09-05

For $abc=1$ prove that $\sum _{\text{cyc}}\frac{1}{a+3}\ge \sum _{\text{cyc}}\frac{a}{{a}^{2}+3}$
I tried TL, BW, the Vasc's Theorems and more, but without success.
I proved this inequality!
I proved also the hardest version: $\sum _{cyc}\frac{1}{a+4}\ge \sum _{cyc}\frac{a}{{a}^{2}+4}$
Thanks all!

Nolan Tyler

BW in the following version does not help.
Let $a={x}^{3}$, $b={y}^{3}$ and $c={z}^{3}$
Hence, we need to prove that
$\sum _{cyc}\frac{1}{{x}^{3}+3xyz}\ge \sum _{cyc}\frac{{x}^{3}}{{x}^{6}+3{x}^{2}{y}^{2}{z}^{2}}$
or
$\sum _{cyc}\frac{1}{{x}^{3}+3xyz}\ge \sum _{cyc}\frac{x}{{x}^{4}+3{y}^{2}{z}^{2}}.$
Now, we can assume that $x=min\left\{x,y,z\right\}$, $y=x+u$ and $z=x+v$
and these substitutions give inequality, which I don't know to prove.
But we can use another BW!
Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that
$\sum _{cyc}\frac{x}{3x+y}\ge \sum _{cyc}\frac{xy}{3{x}^{2}+{y}^{2}}$
or
$\sum _{cyc}\frac{{x}^{3}-{x}^{2}y}{\left(3x+y\right)\left(3{x}^{2}+{y}^{2}\right)}\ge 0.$
Now, let $x=min\left\{x,y,z\right\}$, $y=x+u$ and $z=x+v$
Hence, we need to prove that
$128\left({u}^{2}-uv+{v}^{2}\right){x}^{7}+16\left(16{u}^{3}+23{u}^{2}v-15u{v}^{2}+16{v}^{3}\right){x}^{6}+$
$+32\left(8{u}^{4}+27{u}^{3}v+12{u}^{2}{v}^{2}-11u{v}^{3}+8{v}^{4}\right){x}^{5}+$
$+4\left(32{u}^{5}+193{u}^{4}v+266{u}^{3}{v}^{2}-42{u}^{2}{v}^{3}-33u{v}^{4}+32{v}^{5}\right){x}^{4}+$
$+2\left(8{u}^{6}+178{u}^{5}v+435{u}^{4}{v}^{2}+152{u}^{3}{v}^{3}-99{u}^{2}{v}^{4}+30u{v}^{5}+8{v}^{6}\right){x}^{3}+$
$+uv\left(45{u}^{5}+375{u}^{4}v+291{u}^{3}{v}^{2}-83{u}^{2}{v}^{3}+57u{v}^{4}+3{v}^{5}\right){x}^{2}+$
$+2{u}^{2}{v}^{2}\left(24{u}^{4}+66{u}^{3}v-18{u}^{2}{v}^{2}+13u{v}^{3}+3{v}^{4}\right)x+$
$+{u}^{3}{v}^{3}\left(18{u}^{3}-6{u}^{2}v+3u{v}^{2}+{v}^{3}\right)\ge 0,$
which is obvious.
Done!

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