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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 \(a=b=c=\frac{1}{\sqrt{3}}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\((ab+a+1)^2 \le (a+b+c) \left( a+ a^2b+ \frac 1c \right) = (a+b+c)(a+a^2b+ab)\)
\(\Rightarrow \dfrac{a}{(ab+a+1)^2} \ge \dfrac{a}{(a+b+c)(a+a^2b+ab)}= \dfrac{1}{(a+b+c)(1+ab+b)}\)
Thiết lập các BĐT tương tự rồi cộng theo vế ta có:
\(\sum \dfrac{a}{(ab+a+1)^2} \ge \dfrac{1}{a+b+c} \sum \dfrac{1}{ab+b+1}= \dfrac{1}{a+b+c}\)
c2: Áp dụng BĐT bunyakovsky:
\(\left(a+b+c\right)\left[\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right]\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ca+c+1}\right)^2\)
Xét \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)
\(=\dfrac{ab+a+1}{ab+a+1}=1\)
do đó \(\left(a+b+c\right).VT\ge1\Leftrightarrow VT\ge\dfrac{1}{a+b+c}\)
dấu = xảy ra khi a=b=c=1
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 \(a=b=c=\dfrac{1}{2}\)