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\(A=\dfrac{3}{a}+\dfrac{3a}{4}+\dfrac{9}{2b}+\dfrac{b}{2}+\dfrac{4}{c}+\dfrac{c}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)
Áp dụng bất đẳng thức Cô-si ta được:
\(\dfrac{3}{a}+\dfrac{3a}{4}\ge2\sqrt{\dfrac{3}{a}.\dfrac{3a}{4}}=3\)
\(\dfrac{9}{2b}+\dfrac{b}{2}\ge2\sqrt{\dfrac{9}{2b}.\dfrac{b}{2}}=3\)
\(\dfrac{4}{c}+\dfrac{c}{4}\ge2\sqrt{\dfrac{4}{c}.\dfrac{c}{4}}=2\)
\(\Rightarrow A\ge3+3+2+\dfrac{1}{4}\left(a+2b+3c\right)\)
\(\Rightarrow A\ge8+\dfrac{1}{4}.20=13\)
Vậy Min A=13. Dấu "=" xảy ra \(\Leftrightarrow\) a=2, b=3,c=4
Áp dụng Côsi
\(S=\frac{3}{4}a+\frac{3}{a}+\frac{1}{2}b+\frac{9}{2b}+\frac{1}{4}c+\frac{4}{c}+\frac{1}{4}\left(a+2b+3c\right)\)
\(\ge2\sqrt{\frac{3a}{4}.\frac{3}{a}}+2\sqrt{\frac{b}{2}.\frac{9}{2b}}+2\sqrt{\frac{c}{4}.\frac{4}{c}}+\frac{1}{4}.20\)
\(=3+3+2+5=13\)
Dấu "=" xảy ra khi \(\frac{3a}{4}=\frac{3}{a};\text{ }\frac{b}{2}=\frac{9}{2b};\text{ }\frac{c}{4}=\frac{4}{c};\text{ }a+2b+3c=20\) hay \(a=2;\text{ }b=3;\text{ }c=4\)
Ta có
M = (3a/4+3/a) + ( c/4+4/c) + (b/2+9/2b) + a/4 + b/2 + 3c/4 >= 3 + 2 + 3 +(a+2b+3c)/4 >= 13
Dấu bằng xảy ra khi a=2,b=3,c=4
b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Tương tự cho 2 cái kia rồi cộng lại
\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)
Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)
a+4/a>=2*căn a*4/a=4
b+9/b>=2*căn b*9/b=6
c+16/c>=2*căn c*16/c=8
=>3a/4+b/2+c/4+3/a+9/2b+4/c>=3+3+2=8
a+2b+3c>=20
=>a/4+b/2+3c/4>=5
=>S>=13
Dấu = xảy ra khi a=2; b=3; c=4