Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

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

\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)

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

Bài 3:

\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)

\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

Cho $a=b=c=1$ thì thỏa mãn đẳng thức nhưng $abc+1=2\neq 0$

Bạn xem lại đề. 

a;b;c phải là số dương chứ bạn?

\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\dfrac{b^2\left(a+1\right)}{2b}=a+1-\dfrac{b+ab}{2}\)

Tương tự:

\(\dfrac{b+1}{c^2+1}\ge b+1-\dfrac{c+bc}{2}\) ; \(\dfrac{c+1}{a^2+1}\ge c+1-\dfrac{a+ca}{2}\)

Cộng vế với vế:

\(VT\ge a+b+c+3-\dfrac{1}{2}\left(a+b+c+ab+bc+ca\right)\)

\(VT\ge6-\dfrac{3}{2}-\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{9}{2}-\dfrac{1}{6}\left(a+b+c\right)^2=3=a+b+c\)

\(\Rightarrow VT\ge a+b+c\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Áp dụng BĐT AM-GM ta có:

\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)

\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)

\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)

\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:

\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)

Đẳng thức xảy ra khi \(a=b=c=1\)