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HÌnh như là \(a+b+c\le\dfrac{3}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\)
Áp dụng BĐT Holder ta có:
\(A=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3^3}+\dfrac{1}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)\(\ge\left(3+\dfrac{1}{\dfrac{1}{2}}+\dfrac{1}{\dfrac{1}{2}}\right)^3=343\)
Xảy ra khi \(a=b=c=\dfrac{1}{2}\)
Em bất tài xin phép để link :( Câu hỏi của Anh Khương Vũ Phương - Toán lớp 9 | Học trực tuyến
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a+\dfrac{1}{4a}\text{ ≥}2\sqrt{a.\dfrac{1}{4a}}=2.\dfrac{1}{2}=1\)
\(b+\dfrac{1}{4b}\text{ ≥}2\sqrt{b.\dfrac{1}{4b}}=2.\dfrac{1}{2}=1\)
\(c+\dfrac{1}{4c}\text{ ≥}2\sqrt{c.\dfrac{1}{4c}}=2.\dfrac{1}{2}=1\)
⇒ \(a+b+c+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥}3\)
⇔ \(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{ ≥}3+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥ }3+\dfrac{3}{4}.\dfrac{\left(1+1+1\right)^2}{a+b+c}=3+\dfrac{3}{4}.\dfrac{9}{a+b+c}\text{ ≥}3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{15}{2}\) ⇒ \(A_{MIN}=\dfrac{15}{2}."="\text{⇔}a=b=c=\dfrac{1}{2}\)
Do \(a+b+c=1\) nên Bất đẳng thức trên tương đương:
\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\le\dfrac{3}{4}\)
\(\Leftrightarrow\left(1-\dfrac{1}{1+a}\right)+\left(1-\dfrac{1}{1+b}\right)+\left(1-\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le\dfrac{3}{4}\)
Áp dụng BĐT cauchy-schwarz engel với a;b;c>0 ta có:
\(3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\le3-\dfrac{\left(1+1+1\right)^2}{1+a+1+b+1+c}=3-\dfrac{9}{4}=\dfrac{3}{4}\)
Ta có:
\(\dfrac{a}{2a+b+c}+\dfrac{b}{a+2b+c}+\dfrac{c}{a+b+2c}=\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{4}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{4}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{4}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{4}.\dfrac{\left(1+1\right)^2}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{4}.\dfrac{\left(1+1\right)^2}{\left(a+c\right)+\left(b+c\right)}\)Áp dụng BĐT Cauchy - Schwarz:
\(VT\le\dfrac{a}{4}.\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{b}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)+\dfrac{c}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{1}{4}.3=\dfrac{3}{4}\)\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
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 \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{\left(1-a\right)^2}+\dfrac{1-a}{8}+\dfrac{1-a}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{\left(1-b\right)^2}+\dfrac{1-b}{8}+\dfrac{1-b}{8}\ge\dfrac{3b}{4}\\\dfrac{c^3}{\left(1-c\right)^2}+\dfrac{1-c}{8}+\dfrac{1-c}{8}\ge\dfrac{3c}{4}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{6-2\left(a+b+c\right)}{8}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\)
Vậy \(P_{min}=\dfrac{1}{4}\)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)
a/ \(B=\dfrac{x^2-x-1}{x^2+x+1}\Leftrightarrow Bx^2+Bx+B=x^2-x-1\)
\(\Leftrightarrow\left(B-1\right)x^2+\left(B+1\right)x+B+1=0\)
\(\Delta=\left(B+1\right)^2-4\left(B-1\right)\left(B+1\right)\ge0\)
\(\Leftrightarrow\left(B+1\right)\left(B+1-4B+4\right)\ge0\)
\(\Leftrightarrow\left(B+1\right)\left(5-3B\right)\ge0\)
\(\Rightarrow-1\le B\le\dfrac{5}{3}\) \(\Rightarrow\left[{}\begin{matrix}B_{max}=\dfrac{5}{3}\\B_{min}=-1\end{matrix}\right.\)
b/ Áp dụng BĐT Cô-si:
\(\left\{{}\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\end{matrix}\right.\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
