Giải thích các bước giải:

Ta có:$6a+3b+2c=6\rightarrow \dfrac{a}{1}+\dfrac{b}{2}+\dfrac{c}{3}=1$

Đặt: $x=\dfrac{a}{1}$, $y=\dfrac{b}{2}$, $z=\dfrac{c}{3}$ Suy ra $x+y+z=1$ và $x,y,z>0$

Ta có:

$P=\dfrac{3ab}{\sqrt[]{36a^2+9b^2+8c^2}}+\dfrac{bc}{\sqrt[]{9b^2+4c^2+72a^2}}+\dfrac{2ac}{\sqrt[]{4c^2+36a^2+18b^2}}$

Suy ra:

$P=\dfrac{a.\dfrac{b}{2}}{\sqrt[]{a^2+(\dfrac{b}{2})^2+2.(\dfrac{c}{3})^2}}+\dfrac{\dfrac{b}{2}.\dfrac{c}{3}}{\sqrt[]{(\dfrac{b}{2})2+(\dfrac{c}{3})^2+2a^2}}+\dfrac{a.\dfrac{c}{3}}{\sqrt{(\dfrac{c}{3})^2+a^2+2(\dfrac{b}{2})^2}}$

$\rightarrow P=\dfrac{xy}{\sqrt[]{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt[]{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt[]{z^2+x^2+2y^2}}$

Lại có:

$\dfrac{xy}{\sqrt[]{x^2+y^2+2z^2}}=\dfrac{2xy}{\sqrt[]{(1+1+2)(x^2+y^2+2z^2)}}\le 
\dfrac{2xy}{x+y+2z}=2xy.\dfrac{1}{(x+z)+(y+z)}\\\le 2xy.\dfrac{1}{4}(\dfrac{1}{x+z}+\dfrac{1}{y+z})=\dfrac{1}{2}(\dfrac{xy}{x+z}+\dfrac{xy}{y+z})$

 Chứng minh tương tự ta có:

$\dfrac{yz}{\sqrt[]{y^2+z^2+2x^2}}\le \dfrac{1}{2}.(\dfrac{yz}{y+z}+\dfrac{yz}{x+z})$

$\dfrac{zx}{\sqrt[]{z^2+x^2+2y^2}}\le \dfrac{1}{2}.(\dfrac{zx}{z+y}+\dfrac{zx}{x+y})$

Suy ra: $P\le \dfrac{1}{2}.(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}+\dfrac{yz}{y+z}+\dfrac{yz}{x+z}+\dfrac{zx}{z+y}+\dfrac{zx}{x+y})=\dfrac{1}{2}.(x+y+z)$

$\rightarrow P\le \dfrac{1}{2}$