Dùng bđt Bunhiacopxki

\(\left[a^2+\left(\sqrt{2}b\right)^2+\left(\sqrt{3}c\right)^2\right]\left[1+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right]\ge\left(a+b+c\right)^2=2016^2\)

\(\Rightarrow S\ge\frac{2016^2}{\frac{11}{6}}=\frac{2016^2.6}{11}\)

Dấu bằng xảy ra khi \(\hept{\begin{cases}\frac{a}{1}=\frac{\sqrt{2}b}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3}c}{\frac{1}{\sqrt{3}}}\\a+b+c=2016\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2b=3c\\a+b+c=2016\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{12096}{11}\\b=\frac{6048}{11}\\c=\frac{4032}{11}\end{cases}}\)

Cho mình hỏi, phân thức cuối cùng của câu a phải là \(\frac{1}{c+2a+b}\)chứ

đặt 

\(A=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=>4A=4a+4b+4c+\dfrac{12}{a}+\dfrac{36}{2b}+\dfrac{16}{c}\)

\(=>4A=a+2b+3c+3a+\dfrac{12}{a}+2b+\dfrac{36}{2b}+c+\dfrac{16}{c}\)

áp dụng BDT AM-GM

\(=>\dfrac{12}{a}+3a\ge2\sqrt{12.3}=12\)

\(=>2b+\dfrac{36}{2b}\ge2\sqrt{36}=12\)

\(=>c+\dfrac{16}{c}\ge2\sqrt{16}=8\)

\(=>4A\ge20+12+12+8=52=>A\ge13\)

dấu"=" xảy ra<=>a=2,b=3,c=4

anh nên lên học 24h để được giả đáp tốt hơn !!

Đặt \(\left(a;2b;3c\right)=\left(x;y;z\right)\Rightarrow x+y+z=3\)

\(Q=\dfrac{x+1}{1+y^2}+\dfrac{y+1}{1+z^2}+\dfrac{z+1}{1+x^2}\)

Ta có:

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

Tương tự:

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

Cộng vế:

\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{2}\left(xy+yz+zx\right)\)

\(Q\ge\dfrac{x+y+z}{2}+3-\dfrac{1}{6}\left(x+y+z\right)^2=\dfrac{3}{2}+3-\dfrac{9}{6}=3\)

\(Q_{min}=3\) khi \(x=y=z=1\) hay \(\left(a;b;c\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)