Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)

\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

Ta thấy: \(\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}=\Sigma_{cyc}\frac{a^2+bc}{\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}}\)

Ta lại có: \(\sqrt[3]{\left(a^2b+b^2c\right)\left(bc^2+ca^2\right)\left(c^2a+ab^2\right)}\le\frac{\left(a^2b+b^2c\right)+\left(bc^2+ca^2\right)+\left(c^2a+ab^2\right)}{3}=\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Leftrightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{\Sigma_{cyc}\left(a^2+bc\right)}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{a^2+b^2+c^2+ab+bc+ca}{\frac{1}{3}\Sigma_{cyc}\left(ab\left(a+b\right)\right)}\)

Nhận thấy: \(A=\left(a+b+c\right)\left(a^2+b^2+c^2+ab+bc+ca\right)=a^3+b^3+c^3+3abc+2\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

Theo Schur: \(a^3+b^3+c^3+3abc\ge\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Leftrightarrow A\ge3\Sigma_{cyc}\left(ab\left(a+b\right)\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\frac{a^2+bc}{abc\left(b^2+c^2\right)}}\ge\frac{3\Sigma_{cyc}\left(ab\left(a+b\right)\right)}{\frac{1}{3}\left(a+b+c\right)\Sigma_{cyc}\left(ab\left(a+b\right)\right)}=\frac{9}{a+b+c}\)

đi ,nt ,mình giải cho

nt là gì

3a) ta có \(\frac{a^2}{a+b}=a-\frac{ab}{a+b}>=a-\frac{ab}{2\sqrt{ab}}=a-\frac{\sqrt{ab}}{2}\)

vì \(a,b>0,a+b>=2\sqrt{ab}nên\frac{ab}{a+b}< =\frac{ab}{2\sqrt{ab}}\)

tương tự \(\frac{b^2}{b+c}=b-\frac{bc}{b+c}>=b-\frac{bc}{2\sqrt{bc}}=b-\frac{\sqrt{bc}}{2}\)

tương tự \(\frac{c^2}{c+a}=c-\frac{ca}{c+a}>=c-\frac{ca}{2\sqrt{ca}}=c-\frac{\sqrt{ca}}{2}\)

cộng từng vế BĐT ta được \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=a+b+c-\frac{\sqrt{ab}}{2}-\frac{\sqrt{bc}}{2}-\frac{\sqrt{ca}}{2}=\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}\left(1\right)\)

giả sử \(\frac{2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}}{2}>=\frac{a+b+c}{2}\)

<=> \(2a+2b+2c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=a+b+c\)

<=> \(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca}>=0\)

<=> \(2a+2b+2c-2\sqrt{ab}-2\sqrt{bc}-2\sqrt{ca}>=0\)

<=> \(\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{a}-\sqrt{c}\right)^2>=0\)

(đúng với mọi a,b,c >0) (2)

(1),(2)=> \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}>=\frac{a+b+c}{2}\left(đpcm\right)\)