Đặt \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\) khi đó cần chứng minh:

\(\dfrac{y}{2x+y}+\dfrac{z}{2y+z}+\dfrac{x}{2z+x}\ge1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{y^2}{2xy+y^2}+\dfrac{z^2}{2yz+z^2}+\dfrac{x^2}{2zx+x^2}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2\left(xy+yz+xz\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1=VP\)

Khi \(a=b=c=1\)

a) Dùng (a+b)²≥4ab
Chia hai vế cho a+b ( vì ab khác 0)
Ta có a+b≥\(\frac{4ab}{a+b}\) (Chuyển ab sang a+b) ta có
\(\frac{a+b}{ab}\)≥\(\frac{4}{a+b}\) <=> \(\frac{1}{a}\)+\(\frac{1}{b}\)≥\(\frac{4}{a+b}\)

BĐT cần chứng minh tương đương:

\(\frac{2}{2+a^2b}+\frac{2}{2+b^2c}+\frac{2}{2+c^2a}\ge2\)

\(\Leftrightarrow\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Ta có: \(VT=\sum\frac{a^2b}{1+1+a^2b}\le\frac{1}{3}\sum\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{1}{3}\sum\sqrt[3]{a^4b^2}=\frac{1}{3}\sum\sqrt[3]{a^2.ab.ab}\)

\(VT\le\frac{1}{9}\sum\left(a^2+ab+ab\right)=\frac{1}{9}\left(a+b+c\right)^2=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Đặt \(\left(\frac{1}{2a+1};\frac{1}{2b+1};\frac{1}{2c+1}\right)=\left(x;y;z\right)\Rightarrow x+y+z\ge1\)

Mặt khác do \(a;b;c>0\Rightarrow x;y;z< 1\)

Ta có: \(P=\frac{x}{3-2x}+\frac{y}{3-2y}+\frac{z}{3-2z}\)

Ta có đánh giá: \(\frac{x}{3-2x}\ge\frac{27x-2}{49}\) \(\forall x\in\left(0;1\right)\)

\(\Leftrightarrow9x^2-6x+1\ge0\Leftrightarrow\left(3x-1\right)^2\ge0\) (luôn đúng)

Thiết lập tương tự và cộng lại:

\(P\ge\frac{27\left(x+y+z\right)-6}{49}\ge\frac{21}{49}=\frac{3}{7}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\) hay \(a=b=c=1\)

Áp dụng BĐT AM-GM ta có:

\(\frac{2a}{2a+1}=1-\frac{1}{2a+1}\ge\frac{1}{2b+1}+\frac{1}{2c+1}\)\(\ge2\sqrt{\frac{1}{\left(2b+1\right)\left(2c+1\right)}}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{2b}{2b+1}\ge2\sqrt{\frac{1}{\left(2a+1\right)\left(2c+1\right)}};\frac{2c}{2c+1}\ge2\sqrt{\frac{1}{\left(2a+1\right)\left(2b+1\right)}}\)

Nhân theo vế 3 BĐT trên ta có:

\(\frac{2a}{2a+1}\cdot\frac{2b}{2b+1}\cdot\frac{2c}{2c+1}\ge8\sqrt{\frac{1}{\left(2a+1\right)^2\left(2b+1\right)^2\left(2c+1\right)^2}}\)

\(\Leftrightarrow\frac{8abc}{\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)}\ge\frac{8}{\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)}\)

\(\Leftrightarrow8abc\ge8\Leftrightarrow abc\ge1\) (đúng)

Đẳng thức xảy ra khi \(a=b=c=1\)