Cho a,b,c thoa man a+2b +3c >= 21 cmr
a+b+c +3/a+9/2b+4/c>=13
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D
0
TN
0
AH
Akai Haruma
Giáo viên
5 tháng 3 2017
Bài 1)
Đưa về đồng bậc:
\(\left\{{}\begin{matrix}4x^3-y^3=x+2y\\52x^2-82xy+21y^2=-9\end{matrix}\right.\Rightarrow-9\left(4x^3-y^3\right)=\left(x+2y\right)\left(52x^2-82xy+21y^2\right)\)
\(\Leftrightarrow 8x^3+2x^2y-13xy^2+3y^3=0\)
\(\Leftrightarrow (4x-y)(x-y)(2x+3y)\Rightarrow \) \(\left[{}\begin{matrix}x=y\\4x=y\\2x=-3y\end{matrix}\right.\)
Thay từng TH vào hệ phương trình ban đầu ta thấy chỉ TH \(x=y\) thỏa mãn.
\(\Leftrightarrow (x,y)=(1,1),(-1,-1)\)là nghiệm của HPT
AH
Akai Haruma
Giáo viên
5 tháng 3 2017
Bài 2)
Đặt \(P=a+b+c+\frac{3}{4a}+\frac{9}{8b}+\frac{1}{c}\Rightarrow 4P=4a+4b+4c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\)
\(\Leftrightarrow 4P=(a+2b+3c)+\left(3a+\frac{3}{a}\right)+\left(2b+\frac{9}{2b}\right)+\left(c+\frac{4}{c}\right)\)
Áp dụng bất đẳng thức AM-GM:
\(\left\{{}\begin{matrix}3a+\dfrac{3}{a}\ge6\\2b+\dfrac{9}{2b}\ge6\\c+\dfrac{4}{c}\ge4\end{matrix}\right.\)\(\Rightarrow 4P\geq (a+2b+3c)+6+6+4\geq 10+6+6+4=26\)
\(\Leftrightarrow P\geq \frac{13}{2}\) (đpcm)
Dấu bằng xảy ra khi \((a,b,c)=(1,\frac{3}{2},2)\)
4 tháng 7 2021
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
Điên nhờ...
MV
0
24 tháng 5 2023
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
15 tháng 11 2021
\(A=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\\ A=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\right)\\ A=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{1}{4}\left(a+2b+3c\right)\\ A\ge2\sqrt{\dfrac{3a}{4}\cdot\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}\cdot\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}\cdot\dfrac{4}{c}}+\dfrac{1}{4}\cdot20\\ A\ge3+3+2+5=13\\ A_{min}=13\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
MA
1
24 tháng 1 2023
Sẵn tiện mk chỉ cho bn luôn dạng này nhé.
Phân tích:
Với \(\alpha,\beta,\gamma>0\) thỏa \(\alpha< 2,\beta< 3,\gamma< 4\) ta có:
\(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left[\left(2-\alpha\right)a+\dfrac{3}{a}\right]+\left[\left(3-\beta\right)b+\dfrac{9}{2b}\right]+\left[\left(4-\gamma\right)c+\dfrac{4}{c}\right]+\left(\alpha a+\beta b+\gamma c\right)\)
\(\ge2\sqrt{3.\left(2-\alpha\right)}+2\sqrt{\dfrac{9}{2}.\left(3-\beta\right)}+2\sqrt{4.\left(4-\gamma\right)}+\left(\alpha a+\beta b+\gamma c\right)\)
Chọn \(\alpha,\beta,\gamma\) (thỏa đk trên) sao cho:
\(\left\{{}\begin{matrix}\left(2-\alpha\right)a=\dfrac{3}{a}\\\left(3-\beta\right)b=\dfrac{9}{2b}\\\left(4-\gamma\right)c=\dfrac{4}{c}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{2\left(3-\beta\right)}}\\c=\sqrt{\dfrac{4}{\left(4-\gamma\right)}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{6-4\alpha}}\\c=\sqrt{\dfrac{4}{4-3\alpha}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)
Ta có: \(a+2b+3c\ge20\). Xác định điểm rơi: \(a+2b+3c=20\)
\(\Rightarrow\sqrt{\dfrac{3}{2-\alpha}}+2\sqrt{\dfrac{9}{6-4\alpha}}+3\sqrt{\dfrac{4}{4-3\alpha}}=20\)
Giải ra ta có \(\alpha=\dfrac{5}{4}\Rightarrow\beta=\dfrac{5}{2};\gamma=\dfrac{15}{4}\)
Lời giải:
Ta có: \(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
\(=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{5a}{4}+\dfrac{5b}{2}+\dfrac{15c}{4}\right)\)
\(\ge^{Cauchy}2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(=3+3+2+\dfrac{5}{4}\left(a+2b+3c\right)\)
\(\ge8+\dfrac{5}{4}.20=33\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy \(MinA=33\), đạt được khi \(a=2;b=3;c=4\)
mk chỉnh lại đề nhé: \(a+2b+3c\ge20\)
\(a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\)
\(=\left(\frac{3a}{4}+\frac{3}{a}\right)+\left(\frac{b}{2}+\frac{9}{2b}\right)+\left(\frac{c}{4}+\frac{4}{c}\right)+\left(\frac{a}{4}+\frac{b}{2}+\frac{3c}{4}\right)\)
\(\ge2\sqrt{\frac{3a}{4}.\frac{3}{a}}+2\sqrt{\frac{b}{2}.\frac{9}{2b}}+2\sqrt{\frac{c}{4}.\frac{4}{c}}+\frac{1}{4}\left(a+2b+3c\right)\) (BĐT AM-GM)
\(\ge\)\(3+3+2+\frac{20}{4}=13\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=2,b=3,c=4\)