K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

17 tháng 8 2017

a)Áp dụng BĐT AM-GM ta có

\(\frac{ab\sqrt{ab}}{a+b}\le\frac{ab\sqrt{ab}}{2\sqrt{ab}}=\frac{ab}{2}\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\frac{bc\sqrt{bc}}{b+c}\le\frac{bc}{2};\frac{ac\sqrt{ac}}{a+c}\le\frac{ac}{2}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=Σ\frac{ab\sqrt{ab}}{a+b}\le\frac{ab+bc+ca}{2}=VP\)

Khi \(a=b=c\)

b)Áp dụng tiếp AM-GM:

\(b\sqrt{a-1}\le\frac{b\left(a-1+1\right)}{2}=\frac{ab}{2}\)

\(a\sqrt{b-1}\le\frac{a\left(b-1+1\right)}{2}=\frac{ab}{2}\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=b\sqrt{a-1}+a\sqrt{b-1}\le ab=VP\)

Khi \(a=b=1\)

22 tháng 4 2020

\(VT=\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{1}{\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{2}{\sqrt{c}}}\right)\le\frac{1}{\sqrt{abc}}\Sigma_{cyc}\left(\frac{\sqrt{a}+\sqrt{b}+2\sqrt{c}}{16}\right)=\frac{1}{\sqrt{abc}}\)

Dấu "=" xay ra khi \(a=b=c=\frac{16}{9}\)

NV
20 tháng 6 2020

Đặt vế trái là P và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=4\)

Ta cần chứng minh: \(P=\frac{1}{xy+2yz+zx}+\frac{1}{xy+yz+2zx}+\frac{1}{2xy+yz+zx}\le\frac{1}{xyz}\)

\(P=\frac{1}{xy+yz+yz+zx}+\frac{1}{xy+yz+zx+zx}+\frac{1}{xy+xy+yz+zx}\)

\(P\le\frac{1}{16}\left(\frac{1}{xy}+\frac{2}{yz}+\frac{1}{zx}+\frac{1}{xy}+\frac{1}{yz}+\frac{2}{zx}+\frac{2}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=\frac{1}{4}\left(\frac{x+y+z}{xyz}\right)=\frac{1}{4}.\frac{4}{xyz}=\frac{1}{xyz}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=\frac{4}{3}\) hay \(a=b=c=\frac{16}{9}\)

4 tháng 12 2019

Do \(a+b+c=1\)  nên :

\(VT=\sqrt{\frac{ab}{c\left(a+b+c\right)+ab}}+\sqrt{\frac{bc}{a\left(a+b+c\right)+bc}}+\sqrt{\frac{ca}{b\left(a+b+c\right)+ac}}\)

\(=\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\)

Áp dụng BĐT AM - GM :
\(\sqrt{\frac{ab}{\left(c+a\right)\left(c+b\right)}}\le\frac{1}{2}\left(\frac{a}{c+a}+\frac{b}{c+b}\right)\)

\(\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{c}{c+a}\right)\)

\(\sqrt{\frac{ca}{\left(b+c\right)\left(b+a\right)}}\le\frac{1}{2}\left(\frac{c}{b+c}+\frac{a}{b+a}\right)\)

Cộng theo vế :
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=\frac{1}{3}\)

Chúc bạn học tốt !!!

25 tháng 1 2020

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

25 tháng 1 2020

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

NV
3 tháng 7 2020

4.

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

5.

\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)

Cộng vế với vế:

\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

NV
3 tháng 7 2020

1.

Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)

\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

2.

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng vế với vế:

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

3.

Từ câu b, thay \(c=1\) ta được:

\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)

4 tháng 10 2017

Do a + b + c = 1 nên \(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}=\frac{\sqrt{\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ca\right]}}{\sqrt{c\left(a+b+c\right)+ab}}\)

\(=\frac{\sqrt{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}}{\sqrt{ac+bc+c^2+ab}}=\frac{\sqrt{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(=\sqrt{\left(a+b\right)^2}=a+b\) (1)

Tương tự \(\hept{\begin{cases}\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}=b+c\text{ }\left(2\right)\\\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=a+c\text{ }\left(3\right)\end{cases}}\)

Cộng vế với vế của (1)(2)(3) lại ta được :

\(\frac{\sqrt{\left(a+bc\right)\left(b+ca\right)}}{\sqrt{c+ab}}+\frac{\sqrt{\left(b+ca\right)\left(c+ab\right)}}{\sqrt{a+bc}}+\frac{\sqrt{\left(c+ab\right)\left(a+bc\right)}}{\sqrt{b+ac}}=2\left(a+b+c\right)=2\)

NV
20 tháng 6 2020

\(\sqrt{\frac{ab+2c^2}{1+ab-c^2}}=\sqrt{\frac{ab+2c^2}{a^2+b^2+ab}}=\frac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(a^2+b^2+ab\right)}}\ge\frac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\ge\frac{ab+2c^2}{a^2+b^2+c^2}=ab+2c^2\)

Tương tự: \(\sqrt{\frac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\) ; \(\sqrt{\frac{ca+2b^2}{1+ac-b^2}}\ge ca+2b^2\)

Cộng vế với vế:

\(VT\ge2\left(a^2+b^2+c^2\right)+ab+bc+ca=2+ab+bc+ca\)

13 tháng 5 2021

Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\) \(\left(x,y,z>0\right)\)

Theo đề \(ab+bc+ca=3abc\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{3}{xyz}\)

\(\Rightarrow x+y+z=3\)

Và \(\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)

\(=\sqrt{\frac{\frac{1}{xy}}{\frac{1}{x}+\frac{1}{y}+1}}+\sqrt{\frac{\frac{1}{yz}}{\frac{1}{y}+\frac{1}{z}+1}}+\sqrt{\frac{\frac{1}{zx}}{\frac{1}{z}+\frac{1}{x}+1}}\)

\(=\frac{1}{\sqrt{x+y+xy}}+\frac{1}{\sqrt{y+z+yz}}+\frac{1}{\sqrt{z+x+zx}}\)

\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\) (Cauchy Schwarz)

Ta có: \(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\)

\(=\sqrt{\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2}\)

\(\le\sqrt{3\left(x+y+xy+y+z+yz+z+x+zx\right)}\)

\(=\sqrt{\left[2\left(x+y+z\right)+\left(xy+yz+zx\right)\right]}\)

\(\le\sqrt{6+\frac{\left(x+y+z\right)^2}{3}}=\sqrt{6+\frac{3^2}{3}}=3\)

\(\Rightarrow\sqrt{\frac{ab}{a+b+1}}+\sqrt{\frac{bc}{b+c+1}}+\sqrt{\frac{ca}{c+a+1}}\)

\(\ge\frac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\frac{9}{3}=3\)

Dấu "=" xảy ra khi: \(x=y=z=1\Rightarrow a=b=c=1\)

13 tháng 5 2021

cảm ơn bạn :>