Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Áp dụng BĐT Bunhiacopxki, ta có:
\(\left(a+b+c\right)\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\right)^2\)
Mà \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{bc}{1+bc+1}=1\)
\(\Rightarrow\left(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right)\left(a+b+c\right)\ge1\)
\(\Rightarrow\frac{a}{\left(ab+b+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
ta có \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}=1\)
đặt \(H=\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\)
áp dụng bất đẳng thức bunhiacopxki ta có
\(H\left(a+b+c\right)\ge\left(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\right)^2=1\)
\(\Rightarrow H\ge\frac{1}{a+b+c}\)
hay \(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
Có: \(VT=\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\)
\(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)
Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)được
\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{2\left(ab+bc+ca\right)}\)
Mà\(\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2\ge3\left(ab+bc+ca\right)\)(Chuyển vế đưa thành tổng bình phương)
\(\Rightarrow VT\ge...\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)
Dấu "=" khi a=b=c=1
Ta có : \(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{bc+a.abc}}=\frac{a}{\sqrt{bc+a\left(a+b+c\right)}}\)
\(=\frac{a}{\sqrt{bc+a^2+ab+ac}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)
Áp dụng bđt Cô-si ngược ta có
\(\frac{a}{\sqrt{bc\left(1+a^2\right)}}=\frac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)\)
C/m tương tự được \(\frac{b}{\sqrt{ca\left(1+b^2\right)}}\le\frac{1}{2}\left(\frac{b}{a+b}+\frac{b}{b+c}\right)\)
\(\frac{c}{\sqrt{ab\left(1+c^2\right)}}\le\frac{1}{2}\left(\frac{c}{a+c}+\frac{c}{b+c}\right)\)
Cộng 3 vế của các bđt trên lại ta được
\(A\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{b}{a+b}+\frac{a}{a+c}+\frac{c}{a+c}+\frac{b}{b+c}+\frac{c}{b+c}\right)\)
\(=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b+c=abc\\a=b=c\end{cases}}\Leftrightarrow\hept{\begin{cases}3a=a^3\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a^3-3a=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a\left(a^2-3\right)=0\\a=b=c\end{cases}}\)
\(\Leftrightarrow a=b=c=\sqrt{3}\left(a,b,c>0\right)\)
Vậy \(A_{max}=\frac{3}{2}\Leftrightarrow x=y=z=\sqrt{3}\)
Ta có \(\frac{1}{a+b+1}=\left(1-\frac{1}{b+c+1}\right)+\left(1-\frac{1}{a+c+1}\right)=\frac{b+c}{b+c+1}+\frac{a+c}{a+c+1}\)
\(\ge2\sqrt{\frac{\left(b+c\right)\left(a+c\right)}{\left(b+c+1\right)\left(a+c+1\right)}}\)
Tương tự \(\frac{1}{b+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{\left(a+b+1\right)\left(a+c+1\right)}}\)
\(\frac{1}{a+c+1}\ge2\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{\left(a+b+1\right)\left(b+c+1\right)}}\)
Nhân 3 bđt trên ta có:
\(\frac{1}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\ge\frac{8\left(a+b\right)\left(b+c\right)\left(a+c\right)}{\left(a+b+1\right)\left(b+c+1\right)\left(a+c+1\right)}\)
=> \(\left(a+b\right)\left(b+c\right)\left(a+c\right)\le\frac{1}{8}\)
MaxA=1/8 khi a=b=c=1/4
Áp dụng BĐT Cô-si cho 3 số dương, ta có :
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)
Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)
hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)
\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)
\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Nhân từng vế 2 bất đẳng thức trên, ta được đpcm
Dấu "=" xảy ra khi a = b = c
Vậy ...
2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có :
\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)
Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)
Tương tự : ....
\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)
\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)
Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1