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Lời giải:
Theo hệ quả quen thuộc của BĐT AM-GM thì:
\((a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Leftrightarrow (\sqrt{3})^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 1\)
\(\Rightarrow \frac{a}{\sqrt{a^2+1}}\leq \frac{a}{\sqrt{a^2+ab+bc+ac}}=\frac{a}{\sqrt{(a+b)(a+c)}}\)
Hoàn toàn TT với các phân thức còn lại và cộng theo vế:
\(\Rightarrow \text{VT}\leq \frac{a}{\sqrt{(a+b)(a+c)}}+\frac{b}{\sqrt{(b+c)(b+a)}}+\frac{c}{\sqrt{(c+a)(c+b)}}\)
\(\leq \frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)+\frac{1}{2}\left(\frac{b}{b+c}+\frac{b}{b+a}\right)+\frac{1}{2}\left(\frac{c}{c+a}+\frac{c}{c+b}\right)\) (BĐT Cauchy)
hay \(\text{VT}\leq \frac{1}{2}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{3}{2}\)(đpcm)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Lời giải:
Vì $a+b+c=1$ nên:
\(\text{VT}=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\right)\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{3}{4}\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)+\frac{3}{4}\)
\(=(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab})+(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc})+(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ac})+\frac{3}{4}\)
\(\geq 2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}=\frac{15}{4}\) (áp dụng BĐT AM-GM)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\Rightarrow\dfrac{1}{8}\ge abc\)
Áp dụng BĐT Holder ta có:
\(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3\cdot3\cdot3}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\right)^3\)
\(=\left(3+2\sqrt[3]{\dfrac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\dfrac{1}{\dfrac{1}{8}}}\right)^3=343\)
Khi \(a=b=c=\dfrac{1}{2}\)
\(2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\)
Thay thế \(a+b+c=1\)
\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{2a+b+c}{b+c}+\dfrac{a+2b+c}{a+c}+\dfrac{a+b+2c}{a+b}\)
\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{2a}{b+c}+\dfrac{2b}{a+c}+\dfrac{2c}{a+b}+3\)
\(\Leftrightarrow\dfrac{2b}{a}+\dfrac{2c}{b}+\dfrac{2a}{c}\ge\dfrac{2a}{b+c}+\dfrac{2b}{a+c}+\dfrac{2c}{a+b}+3\)
\(\Leftrightarrow\left(\dfrac{2b}{a}-\dfrac{2b}{a+c}\right)+\left(\dfrac{2c}{b}-\dfrac{2c}{a+b}\right)+\left(\dfrac{2a}{c}-\dfrac{2a}{b+c}\right)\ge3\)
\(\Leftrightarrow\dfrac{2bc}{a\left(a+c\right)}+\dfrac{2ca}{b\left(a+b\right)}+\dfrac{2ab}{c\left(b+c\right)}\ge3\)
\(\Leftrightarrow\dfrac{bc}{a\left(a+c\right)}+\dfrac{ca}{b\left(a+b\right)}+\dfrac{ab}{c\left(b+c\right)}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{3}{2}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c+a+b+c\right)}=\dfrac{\left(ab+bc+ca\right)^2}{2abc}\)
Chứng minh rằng \(\dfrac{\left(ab+bc+ca\right)^2}{2abc}\ge\dfrac{3}{2}\)
\(\Leftrightarrow2\left(ab+bc+ca\right)^2\ge6abc\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\ge3abc\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a^2b^2+b^2c^2\ge2\sqrt{a^2b^4c^2}=2ab^2c\\b^2c^2+c^2a^2\ge2\sqrt{a^2b^2c^4}=2abc^2\\a^2b^2+c^2a^2\ge2\sqrt{a^4b^2c^2}=2a^2bc\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2b^2+b^2c^2+c^2a^2\right)\ge2abc\left(a+b+c\right)\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2\ge abc\) ( đpcm )
Vì \(\dfrac{\left(ab+bc+ca\right)^2}{2abc}\ge\dfrac{3}{2}\)
Vậy \(\dfrac{\left(bc\right)^2}{abc\left(a+c\right)}+\dfrac{\left(ca\right)^2}{abc\left(a+b\right)}+\dfrac{\left(ab\right)^2}{abc\left(b+c\right)}\ge\dfrac{3}{2}\)
\(\Leftrightarrow2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\ge\dfrac{1+a}{1-a}+\dfrac{1+b}{1-b}+\dfrac{1+c}{1-c}\)( đpcm )