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

CMTT: \(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ac}{b+c}+\dfrac{ac}{b+a}\right)\)

\(\Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{ab}{c+a}+\dfrac{ab}{c+b}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ac}{b+c}+\dfrac{ac}{b+c}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left[\dfrac{b\left(a+c\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{c\left(a+b\right)}{a+b}\right]=\dfrac{1}{2}\left(a+b+c\right)=1\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{2}{3}\)

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

Tương tự:

\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\) ; \(\dfrac{ca}{\sqrt{ac+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)

Cộng vế:

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

\(P_{max}=1\) khi \(a=b=c=\dfrac{2}{3}\)

Ta có:

\(\left(2a^2-b^2-c^2\right)^2\ge0\)

\(\Leftrightarrow4a^4+b^4+c^4-4a^2b^2-4a^2c^2+2b^2c^2\ge0\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\ge6a^2b^2+6a^2c^2-3a^4\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge3a^2\left(2b^2+2c^2-a^2\right)\)

\(\Leftrightarrow\dfrac{1}{\sqrt{2b^2+2c^2-a^2}}\ge\dfrac{\sqrt{3}a}{a^2+b^2+c^2}\)

\(\Leftrightarrow\dfrac{a}{\sqrt{2b^2+2c^2-a^2}}\ge\sqrt{3}\dfrac{a^2}{a^2+b^2+c^2}\)

Tương tự: \(\dfrac{b}{\sqrt{2a^2+2c^2-b^2}}\ge\sqrt{3}.\dfrac{b^2}{a^2+b^2+c^2}\) ; \(\dfrac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\sqrt{3}.\dfrac{c^2}{a^2+b^2+c^2}\)

Cộng vế: \(P\ge\dfrac{\sqrt{3}\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=\sqrt{3}\)

\(P_{min}=\sqrt{3}\) khi \(a=b=c\)

Đặt \(\left(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\right)=\left(x,y,z\right)\) với x, y, z > 0 thì ta có \(x+y+z=1\).

Đặt biểu thức ở VT là A. Ta có: 

\(A=\sqrt{\dfrac{b^2+2a^2}{a^2b^2}}+\sqrt{\dfrac{c^2+2b^2}{b^2c^2}}+\sqrt{\dfrac{a^2+2c^2}{c^2a^2}}=\sqrt{x^2+2y^2}+\sqrt{y^2+2z^2}+\sqrt{z^2+2x^2}\).

Ta có bất đẳng thức \(\sqrt{a_1^2+a_2^2}+\sqrt{a_3^2+a_4^2}\ge\sqrt{\left(a_1+a_3\right)^2+\left(a_2+a_4\right)^2}\).

Đây là bđt Mincopxki cho hai bộ số thực và dễ dàng cm bằng biến đổi tương đương.

Do đó \(A\ge\sqrt{\left(x+y\right)^2+\left(\sqrt{2}y+\sqrt{2}z\right)^2}+\sqrt{z^2+2x^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\sqrt{2}y+\sqrt{2}z+\sqrt{2}x\right)^2}=\sqrt{1+2}=\sqrt{3}=VP\).

Đẳng thức xảy ra khi a = b = c = 3.

Vậy...

Tương tự: \(GT\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(VT=\dfrac{\sqrt{a^2+a^2+b^2}}{ab}+\dfrac{\sqrt{b^2+b^2+c^2}}{bc}+\dfrac{\sqrt{c^2+a^2+a^2}}{ca}\)

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

\(VT\ge\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

\(\dfrac{\sqrt{b^2+a^2+a^2}}{ab}\ge\dfrac{\sqrt{\dfrac{1}{3}\left(b+a+a\right)^2}}{ab}=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}\right)\)

Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)

Cộng vế với vế:

\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)

Áp dụng BĐT Cauchy cho 2 số dương:

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

CMTT: \(\sqrt{2b+c}\le\dfrac{2b+c+1}{2},\sqrt{2c+a}\le\dfrac{2c+a+1}{2}\)

\(\Rightarrow T=\sqrt{2a+b}+\sqrt{2b+c}+\sqrt{2c+a}\le\dfrac{2a+b+1+2b+c+1+2c+a+1}{2}=\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)

\(maxT=3\Leftrightarrow2a+b=2b+c=2c+a=1=a+b+c\)

\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

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

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

Tương tự và cộng lại:

\(VT\ge ab+bc+ca+2\left(a^2+b^2+c^2\right)=2+ab+bc+ca\)

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 \(a=b=c\)