Đặt \(\left(a;b;c\right)\rightarrow\left(\frac{2y'z'}{x'^2};\frac{2z'x'}{y'^2};\frac{2x'y'}{z'^2}\right)\) với x', y', z' > 0. Quy về chứng minh:

\(\Sigma_{cyc}\frac{x'^3}{\sqrt{x'^6+8y'^3z'^3}}\ge1\). Đặt \(\left(x'^3;y'^3;z'^3\right)=\left(x;y;z\right)\). Quy về:

\(\Sigma_{cyc}\frac{x}{\sqrt{x^2+8yz}}\ge1\). Đến đây em thấy khá quen thuộc, hình như là bài IMO nào đó, để tối lục lại.

Ok, nó đây: https://olm.vn/hoi-dap/detail/229477332481.html

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 \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{2};-1\right)\)