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Đặt: a + b = x; b + c = y; c + a = z
Thì ta có: x \(\ge\)z \(\ge\)y
Theo đề bài ta có:
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\ge0\)
\(\Leftrightarrow\frac{a-b}{2\left(a+b\right)}+\frac{b-c}{2\left(b+c\right)}+\frac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\frac{z-y}{2x}+\frac{x-z}{2y}+\frac{y-x}{2z}\ge0\)
\(\Leftrightarrow xy^2+yz^2+zx^2-x^2y-y^2z-z^2x\ge0\)
\(\Leftrightarrow\left(y-x\right)\left(z-y\right)\left(z-x\right)\ge0\)(1)
Mà ta lại có
\(\hept{\begin{cases}y-x\le0\\z-x\le0\\z-y\ge0\end{cases}}\)nên (1) đúng
\(\Rightarrow\)ĐPCM
Đấu = xảy ra khi x = y = z hay a = b = c
Đặt b+c=m
a+c=n
a+b=p
=>a+b+c =\(\frac{m+n+p}{2}\)
a=\(\frac{n+p-m}{2}\)
b=\(\frac{m+p-n}{2}\)
c=\(\frac{m+n-p}{2}\)
=>\(\frac{n+p-m}{2m}+\frac{m+n-p}{2n}+\frac{m+n-p}{2p}\)
=\(\frac{1}{2}\left(\frac{n}{m}+\frac{m}{n}\right)\) +\(\frac{1}{2}\left(\frac{p}{m}+\frac{m}{p}\right)\) +\(\frac{1}{2}\left(\frac{p}{n}+\frac{n}{p}\right)\) -\(\frac{3}{2}\) \(\ge\) \(\frac{3}{2}\)
Áp dụng BĐT Cosi cho 2 số \(\frac{n}{m};\frac{m}{n}\) ta được:
Từ chứng minh tiếp ....
\(C=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\ge\frac{3}{2}+1+1+1\)
\(\Leftrightarrow\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\ge\frac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\frac{9}{2}\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow\left[\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9\left(^∗\right)\)
Áp dụng bđt Cauchy :
\(\hept{\begin{cases}\left(b+c\right)+\left(c+a\right)+\left(a+b\right)\ge3\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\\\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\ge3\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}\end{cases}}\)
Nhân vế của các bđt ta được :
\(VT\left(^∗\right)\ge3\sqrt[3]{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\cdot3\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
đặt b + c = x ; c + a = y ; a + b = z
\(\Rightarrow\)a + b + c = \(\frac{x+y+z}{2}\)
\(\Rightarrow a=\frac{y+z-x}{2};b=\frac{x+z-y}{2};c=\frac{x+y-z}{2}\)
\(\Rightarrow C=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(C=\frac{1}{2}.\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}-3\right)\ge\frac{1}{2}\left(6-3\right)=\frac{3}{2}\)
\(VT=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{a^2}{ab+ca}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ca+bc}\ge\left(Schwarz\right)\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà theo Cô-si ta có:
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\) (hằng đẳng thức)
\(\Rightarrow VT\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c
\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Ta c/m BĐT phụ: \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)( b tự c/m nhé. Chuyển vế, c/m VP>=0 là xong )
\(\Rightarrow\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)
\(\Leftrightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
đpcm
a) Áp dụng bất đẳng thức AM-GM ta có ngay :
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2\sqrt{\frac{ab^2c}{ac}}=2\sqrt{b^2}=2\left|b\right|=2b\)( do b > 0 )
=> đpcm
Đẳng thức xảy ra <=> a = b = c
b) Áp dụng bất đẳng thức AM-GM ta có :
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)(1) ( như a) đấy :)) )
tương tự : \(\frac{bc}{a}+\frac{ca}{b}\ge2c\)(2) ; \(\frac{ab}{c}+\frac{ca}{b}\ge2a\)(3)
Cộng (1), (2), (3) theo vế ta có đpcm
Đẳng thức xảy ra <=> a = b = c
c) \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\)
\(=\frac{a^3}{2ab}+\frac{b^3}{2ab}+\frac{b^3}{2bc}+\frac{c^3}{2bc}+\frac{c^3}{2ca}+\frac{a^3}{2ca}\)
\(=\frac{a^2}{2b}+\frac{b^2}{2a}+\frac{b^2}{2c}+\frac{c^2}{2b}+\frac{c^2}{2a}+\frac{a^2}{2c}\)(I)
Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :
\(\left(I\right)\ge\frac{\left(a+b+b+c+c+a\right)^2}{2b+2a+2c+2b+2a+2c}=\frac{\left[2\left(a+b+c\right)\right]^2}{4\left(a+b+c\right)}=\frac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c\)
hay \(\frac{a^3+b^3}{2ab}+\frac{b^3+c^3}{2bc}+\frac{c^3+a^3}{2ca}\ge a+b+c\)(đpcm)
Đẳng thức xảy ra <=> a = b = c
Hy vọng a;b;c dương
Khi đó: \(\frac{a^2}{b^2}+1\ge\frac{2a}{b}\) ; \(\frac{b^2}{c^2}+1\ge\frac{2b}{c}\) ; \(\frac{c^2}{a^2}+1\ge\frac{2c}{a}\)
\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+3\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)\)
\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+3\sqrt[3]{\frac{abc}{abc}}-3\)
\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
Dấu "=" xảy ra khi \(a=b=c\)
Áp dụng BĐT Bunyakovsky dạng phân thức ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\)
\(\ge\frac{\left(a+b+c\right)^2}{ab+bc+bc+ca+ca+ab}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Mà \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{\left(a+b+c\right)^2}{\frac{2\left(a+b+c\right)^2}{3}}=\frac{3}{2}\)
Dấu "=" xảy ra khi: a = b = c
Áp dụng bất đẳng thức \(a^2+b^2\ge2ab\)
ta có\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\frac{ab}{bc}=2\frac{a}{c}\)
tương tự:\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\frac{b}{a}\)
\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{c}{b}\)
Cộng 3 về bất đẳng thức trên lại với nhau ta đươc:\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\right)\)
\(\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)
Dấu "=" xảy ra khi \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)
\(\frac{a}{a+b}\)>= \(\frac{a}{a+a}\)= \(\frac{1}{2}\)( vì a + a >= a + b vì a >= b )
\(\frac{b}{b+c}\) >= \(\frac{b}{b+b}\)= \(\frac{1}{2}\)( vì b + b >= b + c vì b >= c )
\(\frac{c}{c+a}\)>= \(\frac{c}{c+c}\) = \(\frac{1}{2}\)( vì c + c >= c + a vì c>=0 )
Từ 3 điều này suy ra
\(\frac{a}{a+b}\)+ \(\frac{b}{b+c}\)+ \(\frac{c}{c+a}\)>= \(\frac{3}{2}\)
dễ dàng c/m (x+y+z)(1/x+1/y+1/z) \(\ge\) 9,dấu "=" khi x=y=z (*)
a/a+b +b/b+c +c/c+a >= 3/2
<=>(a/b+c + 1) + (b/c+a + 1) + (c/a+b + 1) >= 3/2+1+1+1
<=>(a+b+c)/(b+c) + (a+b+c)/(c+a) + (a+b+c)/(a+b) >= 9/2
<=>2(a+b+c)(1/b+c + 1/c+a + 1/a+b) >= 9/2
<=>[(b+c)+(c+a)+(a+b)](1/b+c + 1/c+a + 1/a+b) >= 9/2 (bđt (*))