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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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Thay abc = 2 vào biểu thức A ta được:
\(A=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc\cdot c}{ac+abc+abc}\\ A=\dfrac{1}{b+1+bc}+\dfrac{b}{bc+b+1}+\dfrac{bc}{1+bc+b}\\ A=\dfrac{1+b+bc}{1+b+bc}\\ A=1\)
Hàng thứ 2 phải sửa lại như vậy:
\(A=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{abc.c}{ac+abc.c+abc}\)
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 \(a=b=c=\dfrac{1}{4}\)
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.
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
Ta có
\(\left\{{}\begin{matrix}\dfrac{3a}{ab+3a+6}=\dfrac{3ac}{abc+3ac+6c}=\dfrac{3ac}{24+3ac+6c}=\dfrac{ac}{8+ac+2c}\\\dfrac{4b}{bc+4b+12}=\dfrac{4ab}{abc+4ab+12a}=\dfrac{4ab}{24+4ab+12a}=\dfrac{ab}{6+ab+3a}=\dfrac{abc}{6c+abc+3ac}=\dfrac{24}{6c+24+3ac}=\dfrac{8}{2c+8+ac}\\\dfrac{2c}{ac+2c+8}\end{matrix}\right.\)
=> \(\dfrac{ac}{ac+2c+8}+\dfrac{2c}{ac+2c+8}+\dfrac{8}{ac+2c+8}=\dfrac{ac+2c+8}{ac+2c+8}=1\)
=>A=1