Kayasari Ryuunosuke Coordinator
14/07/2017 at 13:38-
Lãng Tử Hào Hoa 14/07/2017 at 16:00
We have BĐT equivalent:
\(\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)\) \(\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
Use AM - GM inequality we have:
\(\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{3}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (Thing must prove)
The equality occurs \(\Leftrightarrow a=b=c\)
-
Ace Legona 14/07/2017 at 15:33
Another way - Use Cauchy-Schwarz
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\dfrac{9}{2}\)
By Cauchy-Schwarz: \(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\ge\dfrac{9}{2\left(a+b+c\right)}\)
\(\Rightarrow L.H.S=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\)
\(\ge a+b+c\cdot\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}=R.H.S\)
The equality occurs for \(a=b=c\)
-
Denote \(x=b+c;y=a+c;z=a+b\),then \(x,y,z>0\)
Consider the following expression :
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+1+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}+1\)\(=3+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+\left(\dfrac{y}{z}+\dfrac{z}{y}\right)\)
We have : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}=2\) (AM-GM inequality)
Similarly,\(\dfrac{x}{z}+\dfrac{z}{x}\ge2;\dfrac{y}{z}+\dfrac{z}{y}\ge2\).So,we have :
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3+2+2+2=9\)
\(\Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)
\(\Leftrightarrow2\left(\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\right)\ge9\)
\(\Leftrightarrow1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}\ge\dfrac{9}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)