\(2=\left(a^2+ab+\dfrac{b^2}{4}\right)+\left(a^2-2+\dfrac{1}{a^2}\right)-ab\)

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

\(\Rightarrow ab\ge-2\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(1;-2\right);\left(-1;2\right)\)

Với mọi \(0< a< \dfrac{1}{2}\) ta có:

\(\left(\sqrt{2a}-1\right)^2\ge0\Rightarrow2a+1\ge2\sqrt{2a}\)

\(\Rightarrow1\ge2\sqrt{a}\left(\sqrt{2}-\sqrt{a}\right)\)

\(\Rightarrow\dfrac{1}{\sqrt{2}-\sqrt{a}}\ge2\sqrt{a}\)

Do đó:

\(\dfrac{2+\sqrt{2a}}{2-a}=\dfrac{2-a+a+\sqrt{2a}}{2-a}=1+\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{2}\right)}{\left(\sqrt{2}-\sqrt{a}\right)\left(\sqrt{2}+\sqrt{a}\right)}=1+\dfrac{\sqrt{a}}{\sqrt{2}-\sqrt{a}}\ge1+\sqrt{a}.2\sqrt{a}=2a+1\)

Tương tự:

\(\dfrac{2+\sqrt{2b}}{2-b}\ge2b+1\)

Cộng vế:

\(\dfrac{2+\sqrt{2a}}{2-a}+\dfrac{2+\sqrt{2b}}{2-b}\ge2a+1+2b+1=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)

\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Dấu BĐT bị ngược, sửa đề: \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).

Đặt \(b^2=x\left(x>0\right)\Rightarrow a+x=2ax\).

Khi đó ta cần chứng minh:

\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)

Áp dụng BĐT AM-GM:

\(\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\)

\(\le\dfrac{1}{2a^2x+2ax^2}+\dfrac{1}{2ax^2+2a^2x}\)

\(=\dfrac{2}{2ax\left(a+x\right)}\)

\(=\dfrac{1}{ax\left(a+x\right)}\)

\(=\dfrac{1}{2a^2x^2}\)

Ta thấy: \(a+x\ge2\sqrt{ax}\)

\(\Leftrightarrow2ax\ge2\sqrt{ax}\)

\(\Leftrightarrow ax-\sqrt{ax}\ge0\)

\(\Leftrightarrow\sqrt{ax}\left(\sqrt{ax}-1\right)\ge0\)

\(\Leftrightarrow\sqrt{ax}\ge1\)

\(\Rightarrow ax\ge1\)

Khi đó: \(\dfrac{1}{2a^2x^2}\le\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{a^4+x^2+2ax^2}+\dfrac{1}{a^2+x^4+2a^2x}\le\dfrac{1}{2}\)

Hay \(\dfrac{1}{a^4+b^4+2ab^4}+\dfrac{1}{a^2+b^4+2a^2b^2}\le\dfrac{1}{2}\).

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

BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)

Ta có:

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

\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)

Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)

\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)

Cộng vế với vế:

\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)

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

\(VT=3\left(\dfrac{1}{4ab}+\dfrac{1}{a^2+4b^2}\right)+\dfrac{1}{2.a.2b}\ge\dfrac{12}{a^2+4ab+4b^2}+\dfrac{2}{\left(a+2b\right)^2}=14\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)