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\(\dfrac{a}{\sqrt{b^3+1}}=\dfrac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\ge\dfrac{2a}{b+1+b^2-b+1}=\dfrac{2a}{b^2+2}\)
Tương tự và cộng lại:
\(VT\ge\dfrac{2a}{b^2+2}+\dfrac{2b}{c^2+2}+\dfrac{2c}{a^2+2}=a-\dfrac{ab^2}{b^2+2}+b-\dfrac{bc^2}{c^2+2}+c-\dfrac{ca^2}{a^2+2}\)
\(VT\ge6-\left(\dfrac{ab^2}{b^2+2}+\dfrac{bc^2}{c^2+2}+\dfrac{ca^2}{c^2+2}\right)\)
Ta có:
\(\dfrac{ab^2}{b^2+2}=\dfrac{2ab^2}{2b^2+4}=\dfrac{2ab^2}{b^2+b^2+4}\le\dfrac{2ab^2}{3\sqrt[3]{4b^4}}=\dfrac{a}{3}\sqrt[3]{2b^2}=\dfrac{a}{3}\sqrt[3]{2.b.b}\le\dfrac{a}{9}\left(2+b+b\right)\)
Tương tự và cộng lại:
\(VT\ge6-\left(\dfrac{2a}{9}\left(b+1\right)+\dfrac{2b}{9}\left(c+1\right)+\dfrac{2c}{9}\left(a+1\right)\right)\)
\(=6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{9}\left(ab+bc+ca\right)\ge6-\dfrac{2}{9}\left(a+b+c\right)-\dfrac{2}{27}\left(a+b+c\right)^2=2\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐT
\(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\\ \Rightarrow\dfrac{9abc}{a+3a+2c}\\ =\dfrac{9}{\left(a+c\right)\left(b+c\right)+2b}\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{4}{2}\)
Tương tự với 2 BĐT còn lại rồi cộng vế theo vế
=> 9 vế trái
\(\le\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\\ +\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{a+b+c}{2}\\ =\dfrac{3\left(a+b+c\right)}{2}\\ \Rightarrow......._{\left(đpcm\right)}\)
\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)
((a/b+b/c+c/a)/3)>=\(\sqrt[3]{\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{c}{a}}=1\)
=>a/b+b/c+c/a>=3
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\Leftrightarrow\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c\)\(\Leftrightarrow a+\dfrac{a^2}{b+c}+b+\dfrac{b^2}{c+a}+c+\dfrac{c^2}{a+b}=a+b+c\)
\(\Leftrightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\left(dpcm\right)\)
Lời giải:
Ta có:
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow \left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)(a+b+c)=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{a(b+c)}{b+c}+\frac{b(c+a)}{c+a}+\frac{b^2}{c+a}+\frac{c(a+b)}{a+b}+\frac{c^2}{a+b}=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Leftrightarrow \frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)
Ta có đpcm.
Áp dụng bđt Schwarz ta có: \(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+b+c+c+a}=\dfrac{a+b+c}{2}=1\).
Tham khảo:
\(\dfrac{a+b}{c}+\dfrac{a+c}{b}+\dfrac{b+c}{a}\)
\(=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{a}{b}+\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{c}{a}\)
\(=\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
Áp dụng BĐT cô si, ta có:
\(\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)
\(\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2+2+2=6\left(đpcm\right)\)