Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\Leftrightarrow a=b=c\)
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 .
\(S\le\frac{a}{2a+2b+2c}+\frac{b}{2a+2b+2c}+\frac{c}{2a+2b+2c}=\frac{1}{2}\)
\(S_{max}=\frac{1}{2}\) khi \(a=b=c=1\)
\(P=\frac{ab}{\sqrt{\left(c+a\right)\left(b+c\right)}}+\frac{bc}{\sqrt{\left(c+a\right)\left(a+b\right)}}+\frac{ca}{\sqrt{\left(b+c\right)\left(a+b\right)}}\)
thử dùng cô si đi
Đề thi tuyển sinh chuyên Khoa học tự nhiên-Đại Học quốc gia Hà Nội năm học 2017-2018
ta có: \(ab+bc+ca+abc=2\)
\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)=\left(1+a\right)+\left(1+b\right)+\left(1+c\right)\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)}+\frac{1}{\left(1+b\right)\left(1+c\right)}+\frac{1}{\left(1+c\right)\left(1+a\right)}=1\)
đặt \(x=\frac{1}{1+a};y=\frac{1}{1+b};z=\frac{1}{1+c}\Rightarrow xy+yz+xz=1\)
ta có \(P=\frac{a+1}{\left(a+1\right)^2+1}+\frac{b+1}{\left(b+1\right)^2+1}+\frac{c+1}{\left(c+1\right)^2+1}\)
\(=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}+\frac{\frac{1}{y}}{\frac{1}{y^2}+1}+\frac{\frac{1}{z}}{\frac{1}{z^2}+1}=\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)
\(=\frac{x}{\left(x+y\right)\left(y+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+y\right)\left(z+x\right)}\)
\(=\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}=\frac{2}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)
mà \(9\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left(x+y+z\right)\left(xy+z+zx\right)\)
\(\Leftrightarrow x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\ge6xyz\)(đúng vì theo BĐT Cosi)
\(\Rightarrow P\le\frac{2}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)}=\frac{9}{4\left(x+y+z\right)}\le\frac{9}{4\sqrt{3}}=\frac{3\sqrt{3}}{4}\)
(vì \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3\))
Vậy \(P_{max}=\frac{3\sqrt{3}}{4}\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow a=b=c=\sqrt{3}-1\)
\(\frac{\sqrt{ab}}{a+b+2c}\le\frac{\sqrt{ab}}{2\sqrt{\left(a+c\right)\left(b+c\right)}}\le\frac{\frac{a}{a+c}+\frac{b}{b+c}}{4}\)
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
\(A=\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}\)
\(=\frac{a}{\left(b+a\right)+\left(a+c\right)}+\frac{b}{\left(a+b\right)+\left(c+b\right)}+\frac{c}{\left(a+c\right)+\left(b+c\right)}\)
Áp dụng bđt \(\frac{x}{y+z}\le\frac{1}{4}\left(\frac{x}{y}+\frac{x}{z}\right)\) ta có :
\(A\le\frac{1}{4}\left(\frac{a}{b+a}+\frac{a}{a+c}+\frac{b}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}+\frac{c}{b+c}\right)\)
\(\Leftrightarrow A\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{a+c}{a+c}+\frac{b+c}{b+c}\right)=\frac{1}{4}.3=\frac{3}{4}\) có GTLN là \(\frac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
\(GTLN:\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}=\frac{3}{4}\)