Quy đồng full :)

\(\frac{1}{a\left(1+b\right)}+\frac{1}{b\left(1+c\right)}+\frac{1}{c\left(1+a\right)}\ge\frac{3}{1+abc}\)

\(\Leftrightarrow\frac{1+abc}{a\left(1+b\right)}+\frac{1+abc}{b\left(1+c\right)}+\frac{1+abc}{c\left(1+a\right)}\ge3\)

\(\Leftrightarrow\left[\frac{1+abc}{a\left(1+b\right)}+1\right]+\left[\frac{1+abc}{b\left(1+c\right)}+1\right]+\left[\frac{1+abc}{c\left(1+a\right)}+1\right]\ge6\)

\(\Leftrightarrow\frac{1+abc+ab+a}{a\left(1+b\right)}+\frac{1+abc+bc+b}{b\left(1+c\right)}+\frac{1+abc+c+ac}{c\left(1+a\right)}\ge6\)

\(\Leftrightarrow\frac{ab\left(c+1\right)+\left(a+1\right)}{a\left(1+b\right)}+\frac{bc\left(a+1\right)+\left(b+1\right)}{b\left(1+c\right)}+\frac{ac\left(b+1\right)+\left(c+1\right)}{c\left(1+a\right)}\ge6\)

\(\Leftrightarrow\frac{b\left(c+1\right)}{1+b}+\frac{a+1}{a\left(1+b\right)}+\frac{c\left(a+1\right)}{1+c}+\frac{b+1}{b\left(1+c\right)}+\frac{a\left(b+1\right)}{1+a}+\frac{c+1}{c\left(1+a\right)}\ge6\)

Ta có vế trái tương đương với:

\(\left[\frac{b\left(c+1\right)}{1+b}+\frac{b+1}{b\left(c+1\right)}\right]+\left[\frac{a\left(b+1\right)}{1+a}+\frac{1+a}{a\left(b+1\right)}\right]+\left[\frac{c\left(a+1\right)}{1+c}+\frac{1+c}{c\left(a+1\right)}\right]\)

\(\ge2+2+2=6\)

=> đpcm

\(Q=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\le3-\frac{16}{x+y+z+6}=\frac{1}{3}\)

dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{2};-1\right)\)

Bổ đề: \(\sqrt{u-1}+\sqrt{v-1}\le\sqrt{uv}\left(u,v\ge1\right)\)(*)

Thật vậy: (*)\(\Leftrightarrow u+v-2+2\sqrt{\left(u-1\right)\left(v-1\right)}\le uv\Leftrightarrow\left(u-1\right)\left(v-1\right)+1\ge2\sqrt{\left(u-1\right)\left(v-1\right)}\)(đúng theo bất đẳng thức AM - GM)

Áp dụng bổ đề (*), ta được: \(\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le\sqrt{\left(ab+1\right)-1}+\sqrt{c-1}\le\sqrt{c\left(ab+1\right)}\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}ab\left(c-1\right)=1\\\left(a-1\right)\left(b-1\right)=1\end{cases}}\)

\(S^2=\left(\left|x\right|+\left|y\right|+\left|x\right|\right)^2=x^2+y^2+z^2+2\left(\left|x\right|\left|y\right|+\left|y\right|\left|z\right|+\left|z\right|\left|x\right|\right)\)

\(S^2=x^2+y^2+z^2+\left|x\right|\left(\left|y\right|+\left|z\right|\right)+\left|y\right|\left(\left|z\right|+\left|x\right|\right)+\left|z\right|\left(\left|x\right|+\left|y\right|\right)\)

Áp dụng BĐT chứa dấu GTTĐ ta có:

\(\left|y\right|+\left|z\right|\ge\left|y+z\right|=\left|-x\right|=\left|x\right|\Rightarrow\left|x\right|\left(\left|y\right|+\left|z\right|\right)\ge z^2\)

Cmtt:\(\left|y\right|\left(\left|z\right|+\left|x\right|\right)\ge y^2,\left|z\right|\left(\left|x\right|+\left|y\right|\right)\ge z^2\)

Vì vậy \(S^2\ge2\left(x^2+y^2+z^2\right)\Rightarrow S^2\ge16\Rightarrow S\ge4\)

Dấu "=" xảy ra khi (x;y;z)=(2;-2;0) và hoán vị của nó, ta có S=4