Áp dụng BĐT AM-GM ta có: \(\sqrt{ab}\le\dfrac{a+b}{2}\)
\(\Rightarrow\dfrac{\sqrt{ab}}{2c+a+b}\le\dfrac{\dfrac{a+b}{2}}{\left(a+c\right)+\left(b+c\right)}=\dfrac{a+b}{2\left(a+c\right)}+\dfrac{a+b}{2\left(b+c\right)}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{\sqrt{bc}}{2a+b+c}\le\dfrac{b+c}{2\left(a+b\right)}+\dfrac{b+c}{2\left(a+c\right)};\dfrac{\sqrt{ca}}{2b+c+a}\le\dfrac{c+a}{2\left(b+c\right)}+\dfrac{c+a}{\left(a+b\right)}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{a+2b+c}{2\left(a+c\right)}+\dfrac{a+b+2c}{2\left(a+b\right)}+\dfrac{2a+b+c}{2\left(b+c\right)}\)
\(=\dfrac{a+c}{2\left(a+c\right)}+\dfrac{2b}{2\left(a+c\right)}+\dfrac{a+b}{2\left(a+b\right)}+\dfrac{2c}{2\left(a+b\right)}+\dfrac{b+c}{2\left(b+c\right)}+\dfrac{2a}{2\left(b+c\right)}\)
\(=\dfrac{1}{2}\cdot3+\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
Theo BĐT Nesbitt có: \(VT\le\dfrac{3}{2}+\dfrac{3}{2}=\dfrac{3}{4}=VP\)
e tào lao tý:
giả sử \(a\ge b\ge c\Leftrightarrow2a+b+c\ge2b+a+c\ge2c+a+b\)
\(\Rightarrow\dfrac{\sqrt{bc}}{2a+b+c}\le\dfrac{\sqrt{ac}}{2b+c+a}\le\dfrac{\sqrt{ab}}{2c+a+b}\)
Áp dụng BĐT chebyshev:
\(\left(\dfrac{\sqrt{bc}}{2a+b+c}+\dfrac{\sqrt{ac}}{2b+a+c}+\dfrac{\sqrt{ab}}{2c+a+b}\right)\left(2a+b+c+2b+a+c+2c+a+b\right)\le3\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)\(\Leftrightarrow\left(\dfrac{\sqrt{bc}}{2a+b+c}+\dfrac{\sqrt{ca}}{2b+a+c}+\dfrac{\sqrt{ab}}{2c+a+b}\right).4\left(a+b+c\right)\le3\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
lại có theo AM-GM:\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)
kết hợp với điều trên ta có\(VT\le\dfrac{3}{4}\)
đẳng thức xảy ra khi a=b=c
P/s: Bất đẳng thức chebyshev cho bộ 3 số: \(\left(a+b+c\right)\left(x+y+z\right)\le3\left(ax+by+cz\right)\)nếu \(a\ge b\ge c;x\le y\le z\)
help me
