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Ta có:
\(A=a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\)
\(=\left(\frac{3a}{4}+\frac{3}{a}\right)+\left(\frac{b}{2}+\frac{9}{2b}\right)+\left(\frac{c}{4}+\frac{4}{c}\right)+\left(\frac{a}{4}+\frac{b}{2}+\frac{3c}{4}\right)\)
\(\ge2\sqrt{\frac{3a}{4}.\frac{3}{a}}+2\sqrt{\frac{b}{2}.\frac{9}{2b}}+2\sqrt{\frac{c}{4}.\frac{4}{c}}+\frac{1}{4}.\left(a+2b+3c\right)\)
\(\ge3+3+2+\frac{20}{4}=13\)
Vậy GTNN của A là 13 đạt được khi \(\hept{\begin{cases}a=2\\b=3\\c=4\end{cases}}\)
Lời giải:
\((3a+2b)(3a+2c)=16bc\)
\(\Leftrightarrow 9a^2+6a(b+c)=12bc\)
Theo BĐT Cô-si \(4bc\leq (b+c)^2\Rightarrow 9a^2+6a(b+c)\leq 3(b+c)^2\)
\(\Rightarrow 3a^2+2a(b+c)\leq (b+c)^2\)
\(\Leftrightarrow (b+c)^2-3a^2-2a(b+c)\geq 0\)
\(\Leftrightarrow (b+c)^2-9a^2-2a(b+c)+6a^2\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+3a)-2a(b+c-3a)\geq 0\)
\(\Leftrightarrow (b+c-3a)(b+c+a)\geq 0\)
Vì $a+b+c>0$ nên \(b+c-3a\geq 0\Rightarrow b+c\geq 3a\) (đpcm)
b) Áp dụng BĐT Cô-si và kết quả phần a:
\(\frac{a}{b+c}+\frac{b+c}{a}=\frac{a}{b+c}+\frac{b+c}{9a}+\frac{8(b+c)}{9a}\)
\(\geq 2\sqrt{\frac{a}{b+c}.\frac{b+c}{9a}}+\frac{8(b+c)}{9a}=\frac{2}{3}+\frac{8(b+c)}{9a}\geq \frac{2}{3}+\frac{8.3a}{9a}=\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Ta có đpcm.
Lời giải:
Biến đổi $A$ :
\(A=a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}=\frac{1}{4}(a+2b+3c)+\left(\frac{3a}{4}+\frac{3}{a}\right)+\left (\frac{b}{2}+\frac{9}{2b}\right)+\left (\frac{c}{4}+\frac{4}{c}\right)\)
Ta có: \(\frac{1}{4}(a+2b+3c)\geq \frac{20}{4}=5\)
Áp dụng BĐT AM-GM: \(\left\{\begin{matrix} \frac{3a}{4}+\frac{3}{a}\geq 3\\ \frac{b}{2}+\frac{9}{2b}\geq 3\\ \frac{c}{4}+\frac{4}{c}\geq 2\end{matrix}\right.\)
Do đó \(A\geq 5+3+3+2=13\) hay \(A_{\min}=13\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} a=2\\ b=3\\ c=4\end{matrix}\right.\)
Mấu chốt của bài toán là cách tìm điểm rơi.
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Áp dụng BĐT Cauchy cho 3 số dương a , b , c , ta có :
\(D=\dfrac{a}{a+2b}+\dfrac{b}{b+2c}+\dfrac{c}{c+2a}=\dfrac{a^2}{a^2+2ab}+\dfrac{b^2}{b^2+2bc}+\dfrac{c^2}{c^2+2ac}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Theo BĐT Bu nhi a cốp xki ta có :
\(\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\ge16\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{16}{a+b+c+d}\)
Áp dụng vào bài toán ta có :
\(\dfrac{1}{3a+3b+2c}=\dfrac{1}{16}.\dfrac{16}{\left(a+b\right)+\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3b+3c+2a}=\dfrac{1}{16}.\dfrac{16}{\left(b+c\right)+\left(b+c\right)+\left(a+b\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{b+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3c+3a+2b}=\dfrac{1}{16}.\dfrac{16}{\left(c+a\right)+\left(c+a\right)+\left(a+b\right)+\left(b+c\right)}\le\dfrac{1}{16}\left(\dfrac{1}{c+a}+\dfrac{1}{c+a}+\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{3a+3b+2c}+\dfrac{1}{3b+3c+2a}+\dfrac{1}{3c+3a+2b}\le\dfrac{1}{16}\left(\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\right)=\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}.6=\dfrac{3}{2}\)
Vậy GTLN của A là \(\dfrac{3}{2}\) . Dấu \("="\) xảy ra khi \(a=b=c=\dfrac{1}{4}\)
á mk xl nhá mk ko đọc kĩ đề mk làm nhầm rùi bài mk làm là tìm GTNN nhá bạn ( mất công quá
)
ta có A= a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
= \(\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c}{4}+\dfrac{3c}{4}+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
=\(\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)
vì a,b,c >0 ===> \(\dfrac{3a}{4}>0,\dfrac{3}{a}>0,\dfrac{b}{2}>0,\dfrac{9}{2b}>0,\dfrac{c}{4}>0,\dfrac{4}{c}>0\)
áp dụng BĐT côsi cho các cặp số dương ta đc:
\(\dfrac{3a}{4}+\dfrac{3}{a}>=2.\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}=3\)
\(\dfrac{b}{2}+\dfrac{9}{2b}>=3\)(làm như trên nhá)
\(\dfrac{c}{4}+\dfrac{4}{c}>=2\)
===> \(\dfrac{3a}{4}+\dfrac{3}{a}+\dfrac{b}{2}+\dfrac{9}{2b}+\dfrac{c}{4}+\dfrac{4}{c}>=8\left(1\right)\)
có: \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}=\dfrac{a+2b+3c}{4}\)
mà a+2b+3c >= 20
===> \(\dfrac{a+2b+3c}{4}>=\dfrac{20}{4}=5\)
===> \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}>=5\left(2\right)\)
từ (1) và(2)===> a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}>=13\)
===> A >= 13
Dấu ''='' xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy Min A=13 <=>\(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)