\(Ta có: \(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+2\left(b+c\right)}\) Theo Cauchy: \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)+\left(a+c\right)}+\frac{1}{2\left(b+c\right)}\right)\le\frac{1}{4}\left(\frac{1} {4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{1}{2\left(b+c\right)}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+c\right)}+\frac{1}{b+c}\right)\) Tương tự: \(\frac{1}{3a+2b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+c}\right)\) Và: \(\frac{1}{3a+3b+2c}\le\frac{1}{8}\left(\frac{1}{2\left(a+c\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+b}\right)\) => \(P\le\frac{1}{8}\left(\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}\right)=\frac{1}{4}.2017\) => Pmax  = 2017:4=504,25\)

Ta có: \(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+2\left(b+c\right)}\)

Theo Cauchy: \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

=> \(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)+\left(a+c\right)}+\frac{1}{2\left(b+c\right)}\right)\le\frac{1}{4}\left(\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{1}{2\left(b+c\right)}\right)\)

=> \(\frac{1}{2a+3b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+c\right)}+\frac{1}{b+c}\right)\)

Tương tự: \(\frac{1}{3a+2b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+c}\right)\)

Và: \(\frac{1}{3a+3b+2c}\le\frac{1}{8}\left(\frac{1}{2\left(a+c\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+b}\right)\)

=> \(P\le\frac{1}{8}\left(\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}\right)=\frac{1}{4}.2017\)

=> P_max = 2017:4=504,25

\(Ta có: \(\frac{1}{2a+3b+3c}=\frac{1}{\left(a+b\right)+\left(a+c\right)+2\left(b+c\right)}\) Theo Cauchy: \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)+\left(a+c\right)}+\frac{1}{2\left(b+c\right)}\right)\le\frac{1}{4}\left(\frac{1} {4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{1}{2\left(b+c\right)}\right)\) => \(\frac{1}{2a+3b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(a+c\right)}+\frac{1}{b+c}\right)\) Tương tự: \(\frac{1}{3a+2b+3c}\le\frac{1}{8}\left(\frac{1}{2\left(a+b\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+c}\right)\) Và: \(\frac{1}{3a+3b+2c}\le\frac{1}{8}\left(\frac{1}{2\left(a+c\right)}+\frac{1}{2\left(b+c\right)}+\frac{1}{a+b}\right)\) => \(P\le\frac{1}{8}\left(\frac{2}{a+b}+\frac{2}{a+c}+\frac{2}{b+c}\right)=\frac{1}{4}.2017\) => Pmax  = 2017:4=504,25\)

Ta có \(ab+bc+ca=3abc\)

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

Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) thì ta có \(x,y,z>0;x+y+z=3\) và 

\(\sqrt{\dfrac{a}{3b^2c^2+abc}}=\sqrt{\dfrac{\dfrac{1}{x}}{3.\dfrac{1}{y^2z^2}+\dfrac{1}{xyz}}}=\sqrt{\dfrac{\dfrac{1}{x}}{\dfrac{3x+yz}{xy^2z^2}}}=\sqrt{\dfrac{y^2z^2}{3x+yz}}\) \(=\dfrac{yz}{\sqrt{3x+yz}}\) \(=\dfrac{yz}{\sqrt{x\left(x+y+z\right)+yz}}\) \(=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)

Do đó \(T=\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)

Lại có \(\dfrac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}\)

Lập 2 BĐT tương tự rồi cộng theo vế, ta được \(T\le\dfrac{yz}{2\left(x+y\right)}+\dfrac{yz}{2\left(x+z\right)}+\dfrac{zx}{2\left(y+z\right)}+\dfrac{zx}{2\left(y+x\right)}\) \(+\dfrac{xy}{2\left(z+x\right)}+\dfrac{xy}{2\left(z+y\right)}\)

\(T\le\dfrac{yz+zx}{2\left(x+y\right)}+\dfrac{xy+zx}{2\left(y+z\right)}+\dfrac{xy+yz}{2\left(z+x\right)}\)

\(T\le\dfrac{x+y+z}{2}\) (do \(x+y+z=3\))

\(T\le\dfrac{3}{2}\)

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

Vậy \(maxT=\dfrac{3}{2}\), xảy ra khi \(a=b=c=1\)

 (Mình muốn gửi lời cảm ơn tới bạn Nguyễn Đức Trí vì ý tưởng của bài này chính là bài mình vừa hỏi lúc nãy trên diễn đàn. Cảm ơn bạn Trí rất nhiều vì đã giúp mình có được lời giải này.)

 Bạn Lê Song Phương xem lại dùm nhé, thanks!

\(...\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\)

\(...\Rightarrow T\le2.3=6\)

\(\Rightarrow GTLN\left(T\right)=6\left(tạia=b=c=1\right)\)

 Lúc mình đọc lời giải kia của bạn thì mình thấy cũng hợp lí nhưng mà Cô-si hơi nhầm tí ở chỗ \(\dfrac{1}{z+x}+\dfrac{1}{z+y}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt{\left(z+x\right)\left(z+y\right)}}\) ấy.

 Nên là mình cũng dựa trên ý tưởng của bạn nhưng sửa \(\dfrac{1}{2}\) thành 2 thì mới đúng được

 Không thì bạn cứ kiểm tra bằng cách thay điểm rơi \(a=b=c=1\) vào T thì nó ra \(\dfrac{3}{2}\) ngay chứ không ra 6 đâu.

Bổ đề :\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)

Áp dụng bất đẳng thức Cô-si ta có:

 \(x+y+z\ge3\sqrt[3]{xyz};\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge3\sqrt[3]{\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}}\)

\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}}=9\) 

Dấu "=" xảy ra ⇔ x=y=z

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

Tương tự ta có:\(\dfrac{bc}{b+3c+2a}\le\dfrac{bc}{9}\left(\dfrac{1}{b+a}+\dfrac{1}{c+a}+\dfrac{1}{2c}\right)\)

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

Cộng vế với vế ta có:

\(A\le\dfrac{1}{9}.\left(\dfrac{ab+bc}{a+c}+\dfrac{cb+ac}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(=\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}.\left(6+\dfrac{6}{3}\right)=1\)

Dấu "=" xảy ra ⇔ a=b=c=2

Vậy Max A=1⇔ a=b=c=2

Áp dụng bđt \(\dfrac{9}{a+b+c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

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

Tương tự và cộng theo vế suy ra \(9A\le\dfrac{3\left(a+b+c\right)}{2}=9< =>A\le1\)

Dấu "=" xảy ra khi và chỉ khi a = b = c = 2

\(\left\{{}\begin{matrix}a;b;c\ge0\\a+b+c=1\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le1\)

\(\Rightarrow a\left(a-1\right)\le0\Rightarrow a^2\le a\)

\(\Rightarrow\sqrt{2a^2+3a+4}=\sqrt{a^2+a^2+3a+4}\le\sqrt{a^2+a+3a+4}=a+2\)

Tương tự và cộng lại:

\(\Rightarrow M\le a+2+b+2+c+2=7\)

\(M_{max}=7\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{5}}{2}\left(a+b\right)\)

Tương tự:

\(\sqrt{2b^2+bc+2c^2}\ge\dfrac{\sqrt{5}}{2}\left(b+c\right)\) ; \(\sqrt{2c^2+ca+2a^2}\ge\dfrac{\sqrt{5}}{2}\left(c+a\right)\)

Cộng vế với vế:

\(P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3=\dfrac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{9}\)