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Áp dụng BĐT Cauchy - Schwarz và BĐT phụ \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow M^2=\left(\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{c+a+2b}}+\sqrt{\frac{c}{a+b+2c}}\right)^2\)
\(\le\left(1+1+1\right)\left(\frac{a}{b+c+2a}+\frac{b}{c+a+2b}+\frac{c}{a+b+2c}\right)\)
\(\le\frac{3}{4}\left(\frac{a}{b+a}+\frac{a}{c+a}+\frac{b}{b+c}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{c+b}\right)\)
\(=\frac{3}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{9}{4}\)
\(\Rightarrow M\le\frac{3}{2}\)
Dấu "= " xảy ra \(\Leftrightarrow a=b=c\)
Bìa này muốn làm cân 2 bước nha
Bước 1 ) CM BĐT \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
nó được CM như sau
áp dụng BĐT cô si ta đc
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3.\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=9.\sqrt[3]{xyz.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=9\)
dấu = xảy ra khi x=y=z
Bước 2 ) Theo CM bước 1 . áp dụng ta đc
\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}=\frac{ab}{9}.\frac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}.\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)
CM tương tự ta đc
\(\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{a+c}+\frac{1}{a+b}+\frac{1}{2c}\right)\)
\(\frac{ca}{c+3a+2b}\le\frac{ca}{9}\left(\frac{1}{b+c}+\frac{1}{a+b}+\frac{1}{2a}\right)\)
cộng zế zới zế ta đc
\(A\le\frac{1}{9}\left(\frac{ab+bc}{a+c}+\frac{ab+ca}{b+c}+\frac{bc+ca}{a+b}+\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)\)
\(A\le\frac{1}{9}\left(b+a+c+\frac{a+b+c}{2}\right)=\frac{a+b+c}{6}=\frac{6}{6}=1\)
=> MAx A=1 khi a=b=c=2
Áp dụng bổ đề quen thuộc \(x^3+y^3\ge xy\left(x+y\right)\), ta được: \(\frac{1}{2a^3+b^3+c^3+2}=\frac{1}{\left(a^3+b^3\right)+\left(a^3+c^3\right)+2}\le\frac{1}{ab\left(a+b\right)+ac\left(a+c\right)+2}\)\(=\frac{bc}{ab^2c\left(a+b\right)+abc^2\left(a+c\right)+2bc}=\frac{bc}{b\left(a+b\right)+c\left(a+c\right)+2bc}\)\(\le\frac{bc}{ab+ac+4bc}=\frac{bc}{b\left(a+c\right)+c\left(a+b\right)+2bc}\)\(\le\frac{1}{9}\left(\frac{bc}{b\left(a+c\right)}+\frac{bc}{c\left(a+b\right)}+\frac{bc}{2bc}\right)=\frac{1}{9}\left(\frac{c}{a+c}+\frac{b}{a+b}+\frac{1}{2}\right)\)(1)
Tương tự, ta có: \(\frac{1}{a^3+2b^3+c^3+2}\le\frac{1}{9}\left(\frac{c}{b+c}+\frac{a}{a+b}+\frac{1}{2}\right)\)(2); \(\frac{1}{a^3+b^3+2c^3+2}\le\frac{1}{9}\left(\frac{b}{b+c}+\frac{a}{a+c}+\frac{1}{2}\right)\)(3)
Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(P\le\frac{1}{9}\left(1+1+1+\frac{3}{2}\right)=\frac{1}{2}\)
Vậy giá trị lớn nhất của P là \(\frac{1}{2}\)đạt được khi x = y = z = 1
\(\frac{\sqrt{ab}}{a+b+2c}\le\frac{\sqrt{ab}}{2\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{4}\)
Tương tự cộng lại ta được:
\(F\le\frac{\frac{a}{a+c}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{b}{a+b}}{4}=\frac{3}{4}\)
Dấu "=" xảy ra tại a=b=c
giỏi thì làm bài nÀY nèk
chứ mấy bác cứ đăng linh ta linh tinh lên online math
Linh ta linh tinh gì. ko biết làm thì tôi mới nhờ mọi người chứ
đây là câu cuối bài khảo sat trg tôi. ko làm được thì đừng phát biểu linh tinh
ap dung bdt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\frac{1}{2a+b+c}=\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\)
\(\Rightarrow P\le\frac{1}{16}\left[\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2+\left(\frac{1}{a+b}+\frac{1}{b+c}\right)^2+\left(\frac{1}{b+c}+\frac{1}{a+c}^2\right)\right]\)
\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(a+c^2\right)}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(a+b\right)\left(a+c\right)}\)\(+\frac{2}{\left(b+c\right)\left(c+a\right)}\)
ap dung \(x^2+y^2+z^2\ge xy+yz+xz\) voi a+b=x, b+c=y, c+a=z
\(16P\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)
tiếp tục áp dụng bdt ban đầu \(\frac{4}{a+b}\le\frac{1}{a}+\frac{1}{b}\)
\(\Rightarrow\frac{1}{\left(a+b\right)^2}\le4.16.\left(\frac{1}{a}+\frac{1}{b}\right)^2\)
\(\Rightarrow16P\le\frac{1}{4}.16\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)
=\(\frac{1}{4}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\right)\)
tiep tuc ap dung bo de thu 2 ta co
\(16P\le\frac{1}{4}.4\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)
\(\Rightarrow p\le\frac{3}{16}\)dau =khi a=b=c=1
Ta có :
\(\left(x-y\right)^2\ge0\Rightarrow x^2+y^2\ge2xy\Rightarrow\left(x+y\right)^2\ge4xy\)
\(\Rightarrow\frac{1}{x+y}\le\frac{1}{4}\left(\frac{x+y}{xy}\right)=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Áp dụng BĐT trên ta có :
\(A=\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\)
\(\Rightarrow A=\frac{a}{\left(a+b\right)+\left(a+c\right)}+\frac{b}{\left(a+b\right)+\left(b+c\right)}+\frac{c}{\left(c+a\right)+\left(b+c\right)}\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)+\frac{1}{4}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(+\frac{1}{4}\left(\frac{c}{c+a}+\frac{c}{b+c}\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}+\frac{c}{b+c}\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(\left(\frac{a}{a+b}+\frac{b}{a+b}\right)+\left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+c}+\frac{c}{b+c}\right)\right)\)
\(\Rightarrow A\le\frac{1}{4}\left(1+1+1\right)\)
\(\Rightarrow A\le\frac{3}{4}\)
Dấu " = " xảy ra khi a=b=c
Ta có: \(A=\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\)
\(=\frac{a}{\left(a+b\right)+\left(a+c\right)}+\frac{b}{\left(a+b\right)+\left(b+c\right)}+\frac{c}{\left(a+c\right)+\left(b+c\right)}\)
\(\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)+\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)
\(=\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{a+c}{a+c}\right)=\frac{3}{4}\)
Dấu "=" xảy ra <=> a = b = c
Vậy max A = 3/4 đạt tại a= b = c .